Comprehensive Class 10 Triangles Notes with Examples
Class 10 Triangles Notes cover all essential theorems, properties, and problem-solving techniques, including congruence, similarity, Pythagoras theorem, and properties of medians, altitudes, and angle bisectors. Each concept is explained with examples and practical applications for easier understanding. These notes help students learn step-by-step methods to solve triangle-related problems efficiently. Understanding triangles is fundamental for geometry, trigonometry, and real-life applications in construction and design. The notes also include tips for quick identification of triangle properties, shortcuts for solving congruence and similarity questions, and exam-focused practice problems. By following these notes, students can strengthen their reasoning and visualization skills, improve accuracy, and gain confidence in solving triangle-based questions in Class 10 board exams and competitive tests. Solving the NCERT questions with the help of the NCERT Solutions for class 10 Maths will help in building the foundation
Class 10 Triangles Notes, Solved example & Questions
Class 10 Triangles Notes cover all essential theorems, properties, and problem-solving techniques, including congruence, similarity, Pythagoras theorem, and properties of medians, altitudes, and angle bisectors. Each concept is explained with examples and practical applications for easier understanding. These notes help students learn step-by-step methods to solve triangle-related problems efficiently.
Understanding triangles is fundamental for geometry, trigonometry, and real-life applications in construction and design. The notes also include tips for quick identification of triangle properties, shortcuts for solving congruence and similarity questions, and exam-focused practice problems. By following these notes, students can strengthen their reasoning and visualization skills, improve accuracy, and gain confidence in solving triangle-based questions in Class 10 board exams and competitive tests. Solving the NCERT questions with the help of the NCERT Solutions for class 10 Maths will help in building the foundation
CONGRUENT AND SIMILAR FIGURES
Two geometric figures having the same shape and size are known as congruent figures. Geometric figures having the same shape but different sizes are known as similar figures.
SIMILAR TRIANGLES:
Two triangles ABC and DEF are said to be similar if their:
(i) Corresponding angles are equal.
i.e. ∠A = ∠D, ∠B = ∠E, ∠C = ∠F
And,
(ii) Corresponding sides are proportional i.e. AB/DE = BC/EF = AC/DF.
CHARACTERISTIC PROPERTIES OF SIMILAR TRIANGLES:
(i) (AAA Similarity) If two triangles are equiangular, then they are similar.
(ii) (SSS Similarity) If the corresponding sides of two triangles are proportional, then they are similar.
(iii) (SAS Similarity) If in two triangle's one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.
RESULTS BASED UPON CHARACTERISTIC PROPERTIES OF SIMILAR TRIANGLES:
(i) If two triangles are equiangular, then the ratio of the corresponding sides is the same as the ratio of the corresponding medians.
(ii) If two triangles are equiangular, then the ratio of the corresponding sides is same at the ratio of the corresponding angle bisector segments.
(iii) If two triangles are equiangular then the ratio of the corresponding sides is same at the ratio of the corresponding altitudes.
(iv) If one angle of a triangle is equal to one angle of another triangle and the bisectors of these equal angles divide the opposite side in the same ratio, then the triangles are similar.
(v) If two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of another triangle, then the triangles are similar.
(vi) If two sides and a median bisecting the third side of a triangle are respectively proportional to the corresponding sides and the median another triangle, then two triangles are similar.
BASIC PROPORTIONALITY THEOREM (THALES THEOREM)
THEOREM 1:
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.4
To prove: AD/DB = AE/EC
Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB.
Proof: area of ∠ADE
(Taking AD as base)
So, ar(ADE) = 1/2 × AD × EN [The area of ∠ADE is denoted as ar(ADE)].
Similarly, ar(BDE) = 1/2 × DB × EN
and ar(ADE) = 1/2 × AE × DM (Taking AE as base)
Therefore, ar(ADE)/ar(BDE) = AD/DB ...(i)
and ar(ADE)/ar(DEC) = AE/EC ...(ii)
ar(BDE) = ar(DEC) ... (iii)
[∠BDE and DEC are on the same base DE and between the same parallels BC and DE.]
Therefore, from (i), (ii) and (iii), we have: AD/DB = AE/EC.
Corollary: From above equation we have AD/DB = AE/EC.
Adding '1' to both sides we have
AD/DB + 1 = AE/EC + 1
⇒ (AD + DB)/DB = (AE + EC)/EC
⇒ AB/DB = AC/EC.
Theorem 2 - Converse of BPT
THEOREM 2:
(Converse of BPT theorem)
Statement: If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side.
Given: In ∠ABC, DE is a straight line such that AD/DB = AE/EC.
To prove: DE || BC.
Construction: If DE is not parallel to BC, draw DF meeting AC at F.
Proof:
In ∠ABC, let DF || BC
AD/DB = AF/FC ...(i)
[A line drawn parallel to one side of a ∠ divides the other two sides in the same ratio.]
But AD/DB = AE/EC ...(ii) [given]
From (i) and (ii), we get
AF/FC = AE/EC.
Adding 1 to both sides, we get
(AF + FC)/FC = (AE + EC)/EC
AC/FC = AC/EC
FC = EC.
It is possible only when E and F coincide
Hence, DE || BC.
SOME IMPORTANT RESULTS AND THEOREMS:
- The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
- In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC, then AD is the bisector of ∠A.
- The external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle.
- The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.
- The line joining the mid-points of two sides of a triangle is parallel to the third side.
- The diagonals of a trapezium divide each other proportionally.
- If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.
- Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.
- If three or more parallel lines are intersected by two transversal, then the intercepts made by them on the transversal are proportional.
SOLVED PROBLEMS
Problem 1:
Question: In a ∠ABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.
Solution:
In ∠ABC, we have
DE||BC.
AD/BD = AE/CE. [By Basic Proportionality Theorem]
(4x - 3)/(3x - 1) = (8x - 7)/(5x - 3)
(4x – 3)(5x – 3) = (8x – 7)(3x – 1)
20x² – 15x – 12x + 9 = 24x² – 21x – 8x + 7
20x² – 27x + 9 = 24x² – 29x + 7
4x² – 2x – 2 = 0
2x² – x – 1 = 0
(2x + 1)(x – 1) = 0
x = 1 or x = –1/2.
So, the required value of x is 1.
[x = -1/2 is neglected as length cannot be negative].
Problem 2:
Question: D and E are respectively the points on the sides AB and AC of a ∠ABC such that AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm, show that DE || BC.
Solution:
We have,
AB = 12 cm, AC = 18 cm, AD = 8 cm and AE = 12 cm.
BD = AB - AD = (12 – 8) cm = 4 cm.
CE = AC – AE = (18 – 12) cm = 6 cm.
Now, AD/BD = 8/4 = 2.
And, AE/CE = 12/6 = 2
AD/BD = AE/CE.
Thus, DE divides sides AB and AC of ∠ABC in the same ratio. Therefore, by the converse of basic proportionality theorem we have DE||BC.
Problem 3:
Question: In a trapezium ABCD AB||DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such that BE/EC = 4/3. Diagonal DB intersects EF at G. Prove that 7FE = 10AB.
Solution:
In ∠DFG and ∠DAB,
∠1 = ∠2 [Corresponding ∠s ∵ AB || FG]
∠FDG = ∠ADB [Common]
∠DFG ~ ∠DAB [By AA rule of similarity]
FG/AB = DF/DA .....(i)
Again in trapezium ABCD
EF||AB||DC
BE/EC = AF/FD
4/3 = AF/FD
FD/AF = 3/4
(FD + AF)/AF = (3 + 4)/4
DA/AF = 7/4
AF/DA = 4/7
DF/DA = (DA - AF)/DA = 1 - 4/7 = 3/7 …….(ii)
From (i) and (ii), we get
FG/AB = 3/7 i.e. FG = (3/7)AB ......(iii)
In ∠BEG and ∠BCD, we have
∠BEG = ∠BCD [Corresponding angle ∵ EG||CD]
∠GBE = ∠DBC [Common]
∠BEG ~ ∠BCD [By AA rule of similarity]
EG/CD = BE/BC
EG/CD = BE/(BE + EC)
EG/CD = 4/(4 + 3) = 4/7
EG = (4/7)CD = (4/7) × 2AB = (8/7)AB .....(iv)
Adding (iii) and (iv), we get
FG + EG = (3/7)AB + (8/7)AB
FE = (11/7)AB
7FE = 10AB. Hence proved.
Problem 4:
Question: In ∠ABC, if AD is the bisector of ∠A, prove that AB/AC = BD/DC.
Solution:
In ∠ABC, AD is the bisector of ∠A.
BD/DC = AB/AC ....(i) [By internal bisector theorem]
From A draw AL ⊥ BC
AB/AC = BD/DC. [From (i)] Hence Proved.
Problem 5:
Question: ∠BAC = 90°, AD is its bisector. If DE ⊥ AC, prove that DE × (AB + AC) = AB × AC.
Solution:
It is given that AD is the bisector of ∠A of ∠ABC.
BD/DC = AB/AC
(BD + DC)/DC = (AB + AC)/AC [Adding 1 on both sides]
BC/DC = (AB + AC)/AC
DC/BC = AC/(AB + AC). ...(i)
In ∠'s CDE and CBA, we have
∠DCE = ∠BCA [Common]
∠DEC = ∠BAC. [Each equal to 90°]
So by AA-criterion of similarity
∠CDE ~ ∠CBA
DE/AB = DC/BC ...(ii)
From (i) and (ii), we have
DE/AB = AC/(AB + AC)
DE × (AB + AC) = AB × AC.
Problem 6:
Question: In the given figure, PA, QB and RC are each perpendicular to AC. Prove that 1/x + 1/z = 1/y.
Solution:
In ∠PAC, we have BQ||AP
∠CBQ ~ ∠CAP
BQ/AP = BC/AC
y/x = (AC - AB)/AC = 1 - AB/AC. ...(i)
In ∠ACR, we have BQ||CR
∠ABQ ~ ∠ACR
BQ/CR = AB/AC
y/z = AB/AC. ...(ii)
Adding (i) and (ii), we get
y/x + y/z = 1 - AB/AC + AB/AC
y/x + y/z = 1
1/x + 1/z = 1/y. Hence Proved
Problem 7:
Question: In the given figure, AB||CD. Find the value of x.
Solution:
Since the diagonals of a trapezium divide each other proportionally.
(x - 3)/(x - 4) = (3x - 19)/(x - 4)
(x - 3)(x - 4) = (3x - 19)(x - 4)
12x – 76 = x² – 4x – 3x + 12
x² – 19x + 88 = 0
x² – 11x – 8x + 88 = 0
(x – 8)(x – 11) = 0
x = 8 or x = 11.
CRITERIA FOR SIMILARITY OF TWO TRIANGLES
Two triangles are said to be similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportional).
Thus, two triangles ABC and DEF are similar if
(i) ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and
(ii) AB/DE = BC/EF = AC/DF.
In this section, we shall make use of the theorems discussed in earlier sections to derive some criteria for similar triangles which in turn will imply that either of the above two conditions can be used to define the similarity of two triangles.
AAA and SSS Similarity Theorems
CHARACTERISTIC PROPERTY 1 (AAA SIMILARITY)
THEOREM 3:
Statement: If in two triangles, the corresponding angles are equal, then the triangles are similar.
Given: Two triangles ABC and DEF in which ∠A = ∠D, ∠B = ∠E, ∠C = ∠F.
To prove: ∠ABC ~ ∠DEF.
Proof:
Case 1: When AB = DE
In triangles ABC and DEF, we have
∠A = ∠D [Given]
AB = DE [Given]
∠B = ∠E [Given]
∴ ∠ABC ≅ ∠DEF [By ASA congruency]
⇒ BC = EF and AC = DF [c.p.c.t.]
Thus AB/DE = BC/EF = AC/DF [corresponding sides of similar ∠s are proportional]
Hence ∠ABC ~ ∠DEF
Case 2: When AB < DE
Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.
In ∠ABC and ∠DPQ, we have
AB = DP [By construction]
∠A = ∠D [Given]
AC = DQ [By construction]
∴ ∠ABC ≅ ∠DPQ [By SAS congruency]
∴ ∠ABC = ∠DPQ [c.p.c.t.] …(i)
But ∠ABC = ∠DEF [Given] …(ii)
∴ ∠DPQ = ∠DEF [c.p.c.t.]
But ∠DPQ and ∠DEF are corresponding angles.
⇒ PQ || EF
∴ DP/DE = DQ/DF [Corollary to BPT Theorem]
∴ AB/DE = AC/DF. [∵ DP = AB and DQ = AC (by construction)]
Similarly AB/DE = BC/EF.
∴ AB/DE = BC/EF = AC/DF.
Hence, ∠ABC ~ ∠DEF.
Case 3: When AB > DE
Let P and Q be points on AB and AC respectively such that AP = DE and AQ = DF. Join PQ.
In ∠APQ and ∠DEF, we have
AP = DE [By construction]
AQ = DF [By construction]
∠A = ∠D [Given]
∴ ∠APQ ≅ ∠DEF [By SAS congruency]
∴ ∠APQ = ∠DEF [c.p.c.t.] …(i)
But ∠DEF = ∠ABC. [Given] …(ii)
From (i) and (ii) we have
∴ ∠APQ = ∠ABC.
But ∠APQ and ∠ABC are corresponding angles
∴ PQ || BC
∴ AP/AB = AQ/AC [Corollary to BPT Theorem]
∴ DE/AB = DF/AC [∵ AP = DE and AQ = DF (by construction)]
Similarly DE/AB = EF/BC.
Thus DE/AB = EF/BC = DF/AC
Or, AB/DE = BC/EF = AC/DF.
Hence ∠ABC ~ ∠DEF.
CHARACTERISTIC PROPERTY 2 (SSS SIMILARITY)
THEOREM 4:
Statement: If the corresponding sides of two triangles are proportional, then they are similar.
Given: Two triangles ABC and DEF such that
AB/DE = BC/EF = CA/FD
To prove: ∠ABC ~ ∠DEF
Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ
Proof:
AB/DE = AC/DF [Given]
⇒ DP/DE = DQ/DF …(i)
[∵ DP = AB and DQ = AC (by construction)].
In ∠DEF, we have
DP/DE = DQ/DF [From (i)]
∴ PQ || EF [By the converse of BPT]
∴ ∠DPQ = ∠DEF and ∠DQP = ∠DFE [Corresponding angles]
∴ ∠DPQ ~ ∠DEF [By AA similarity] …(ii)
∴ DP/DE = PQ/EF
or AB/DE = PQ/EF [∵ DP = AB] …(iii)
But AB/DE = BC/EF [given] …(iv)
From equations (iii) and (iv), we have
PQ/EF = BC/EF
⇒ BC = PQ …(v)
In ∠ABC and ∠DPQ, we have
AB = DP [By construction]
AC = DQ [By construction]
BC = PQ [By (v)]
∴ ∠ABC ≅ ∠DPQ [by SSS congruency]
⇒ ∠ABC ~ ∠DPQ. [∵ ∠ABC ≅ ∠DPQ ⇔ ∠ABC ~ ∠DPQ] ... (vi)
From equation (ii) and (vi), we get
∠ABC ~ ∠DEF.
CHARACTERISTIC PROPERTY 3 (SAS SIMILARITY)
THEOREM 5:
Statement: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional than the two triangles are similar.
Given: Two triangle ABC and DEF such that ∠A = ∠D and AB/DE = AC/DF.
To prove: ∠ABC ~ ∠DEF.
Construction: Let P and Q be points on DE and DF respectively such that DP = AB and DQ = AC. Join PQ.
Proof:
In ∠ABC and ∠DPQ, we have
AB = DP [By construction]
AC = DQ [By construction]
∠A = ∠D [Given]
∠ABC ≅ ∠DPQ [By SAS congruency] …(i)
Now, AB/DE = AC/DF …(ii)
DP/DE = DQ/DF. [∵ AB = DP and AC = DQ (by construction)]
In ∠DEF, we have
DP/DE = DQ/DF. [From (ii)]
PQ || EF [By the converse of BPT]
∠DPQ = ∠DEF and ∠DQP = ∠DFE [Corresponding angles]
∠DPQ ~ ∠DEF [By AA similarity] …(iii)
From equations (i) and (iii), we get ∠ABC ~ ∠DEF.
SOLVED PROBLEMS
Problem 1:
Question: Examine each pair of triangles in figure and state which pair of triangles are similar. Also, state the similarity criterion used and write the similarity relation in symbolic form.
Solution:
(i) In triangles ABC and PQR, we observe that
∠A = ∠Q = 40°, ∠B = ∠P = 60° and ∠C = ∠R = 80°
Therefore, by AAA-criterion of similarity
∠BAC ~ ∠PQR.
(ii) In triangle PQR and DEF, we observe that
PQ/DE = QR/EF = PR/DF = 2/1 = 2.
Therefore, by SSS-criterion of similarity, we have
∠PQR ~ ∠DEF
(iii) In ∠'s MNP and EFG, we observe that
MN/EF = 2.5/5 = 1/2, NP/FG = 5/10 = 1/2, MP/EG = 6/8 = 3/4.
Therefore, these two triangles are not similar as they do not fulfill SSS-criterion of similarity.
(iv) In ∠'s DEF and MNP, we have
∠D = ∠M = 70°
∠E = ∠N = 80°. [∵ ∠N = 180° – ∠M – ∠P = 180° – 70° – 30° = 80°]
So by AA-criterion of similarity ∠DEF ~ ∠MNP.
Problem 2:
Question: Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Solution:
Given: ∠ABC and ∠PQR in which
BC = a, CA = b, AB = c and QR = p, RP = q, PQ = r.
Also ∠ABC ~ ∠PQR.
To Prove: (a + b + c)/(p + q + r) = a/p = b/q = c/r.
Proof:
Since ∠ABC and ∠PQR are similar, therefore their corresponding sides are proportional.
Let a/p = b/q = c/r = k …(i)
a = kp, b = kq and c = kr
(a + b + c)/(p + q + r) = (kp + kq + kr)/(p + q + r)
= k(p + q + r)/(p + q + r) = k. ... (ii)
From (i) and (ii), we get
(a + b + c)/(p + q + r) = a/p = b/q = c/r. [each equal to k]
Problem 3:
Question: A vertical stick 12m long casts a shadow 8m long on the ground. At the same time a tower casts the shadow 40m long on the ground. Determine the height of the tower.
Solution:
Let AB be the vertical stick and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Join BC and EF.
Let DE = x metres.
We have,
AB = 12m, AC = 8m, and DF = 40m.
In ∠ABC and ∠DEF, we have
∠A = ∠D = 90° and ∠C = ∠F. [Angle of elevation of the sun]
Therefore, by AA-criterion of similarity
∠ABC ~ ∠DEF
AB/DE = AC/DF
12/x = 8/40
x = (12 × 40)/8 = 60 metres.
Problem 4:
Question: In the given figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∠ABD ~ ∠ECF.
Solution:
Given: An isosceles ∠ABC in which AB = AC, E is a point on CB produced, AD ⊥ BC and EF ⊥ AC.
To prove: ∠ABD ~ ∠ECF.
Proof:
In ∠ABC, since AB = AC, therefore ∠C = ∠B [∠s opposite to equal sides are equal]
In ∠ABD and ∠ECF
∠B = ∠C [Proved above]
and ∠EFC = ∠ADB [each 90°]
∠ABD ~ ∠ECF. [AA similarity]
Problem 5:
Question: In figure, if ∠ABE ≅ ∠ACD, prove that ∠ADE ~ ∠ABC.
Solution:
Given: ∠ABE ≅ ∠ACD.
To prove: ∠ADE ~ ∠ABC.
Proof:
Since ∠ABE ≅ ∠ACD
AB = AC [cpct] .... (i)
and, AD = AE .... (ii)
Also, AB/AC = AD/AE [From (i) and (ii)]
AB/AC = AD/AE = 1. … (iii)
Thus, in triangles ADE and ABC, we have
AD/AB = AE/AC and, ∠BAC = ∠DAE. [Common]
Hence, by SAS criterion of similarity.
∠ADE ~ ∠ABC.
Problem 6:
Question: In figure, if BD ⊥ AC and CE ⊥ AB, prove that
(i) ∠AEC ~ ∠ADB.
(ii) CA/AB = CE/DB.
Solution:
Given: BD ⊥ AC and CE ⊥ AB.
To prove: (i) ∠AEC ~ ∠ADB. (ii) CA/AB = CE/DB.
Proof:
(i) In ∠'s AEC and ADB, we have ∠AEC = ∠ADB = 90° [∵ CE ⊥ AB and BD ⊥ AC]
and, ∠EAC = ∠DAB [Each equal to ∠A]
∠AEC ~ ∠ADB. [By AA similarity]
(ii) We have,
∠AEC ~ ∠ADB [Proved above]
CA/AB = CE/DB
CA/AB = CE/DB.
Problem 7:
Question: If CD and GH (D and H lie on AB and FE) are respectively bisectors of ∠ACB and ∠EGF and ∠ABC ~ ∠FEG, prove that
(i) CD/AC = GH/FG.
(ii) ∠DCB ~ ∠HGE.
Solution:
Given:
(i) ∠ABC ~ ∠FEG.
(ii) CD, the bisector of ∠ACB meets AB at D and
(iii) GH, the bisector of ∠EGF meets FE at H.
To prove: (i) CD/AC = GH/FG. (ii) ∠DCB ~ ∠HGE.
Proof:
(i) ∠ABC ~ ∠FEG [Given]
∠BAC = ∠EFG …(i)
∠ABC = ∠FEG …(ii)
and ∠ACB = ∠FGE …(iii)
[Corresponding angles of similar triangles]
(1/2)∠ACB = (1/2)∠FGE.
∠ACD = ∠FGH. …(iv)
[As CD is the bisector of ∠ACB]
and ∠DCB = ∠HGE. …(v)
[As GH is the bisector of ∠EGF]
In ∠ACD and ∠FGH, we have
∠DAC = ∠HFG [From (i)]
∠ACD = ∠FGH [From (iv)]
∠ACD ~ ∠FGH [By AA similarity]
CD/AC = GH/FG.
(ii) In ∠DCB and ∠HGE, we have
∠DBC = ∠HEG [From (ii)]
∠DCB = ∠HGE [From (v)]
∠DCB ~ ∠HGE. [By AA similarity]
AREAS OF SIMILAR TRIANGLES
STATEMENT:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given: Two triangles ABC and PQR such that ∠ABC ~ ∠PQR [Shown in the figure]
To Prove: ar(ABC)/ar(PQR) = AB²/PQ² = BC²/QR² = AC²/PR².
Construction: Draw altitudes AM and PN of the triangle ABC an PQR.
Proof:
ar(ABC) = (1/2) BC × AM.
And ar(PQR) = (1/2) QR × PN.
So, ar(ABC)/ar(PQR) = (BC × AM)/(QR × PN) = (BC/QR) × (AM/PN) ....(i)
Now, in ∠ABM and ∠PQN,
And ∠B = ∠Q [As ∠ABC ~ ∠PQR]
∠M = ∠N. [90° each]
So, ∠ABM ~ ∠PQN. [AA similarity criterion]
Therefore, AM/PN = AB/PQ ....(ii)
Also, ∠ABC ~ ∠PQR. [Given]
So, AB/PQ = BC/QR = AC/PR. .....(iii)
Therefore, ar(ABC)/ar(PQR) = (BC/QR) × (AM/PN) [From (i) and (ii)]
= (BC/QR) × (AB/PQ) [From (iii)]
= (AB/PQ) × (AB/PQ) = AB²/PQ².
Now using (iii), we get
ar(ABC)/ar(PQR) = AB²/PQ² = BC²/QR² = AC²/PR²
PROPERTIES OF AREAS OF SIMILAR TRIANGLES:
- The areas of two similar triangles are in the ratio of the squares of corresponding altitudes.
- The areas of two similar triangles are in the ratio of the squares of the corresponding medians.
- The area of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments.
SOLVED PROBLEMS ON AREAS
Problem 1:
Question: Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on this diagonals.
Solution:
Given: A square ABCD. Equilateral triangles BCE and ACF have been described on side BC and diagonals AC respectively.
To prove: Area (BCE) = (1/2) Area (ACF).
Proof:
Since ∠BCE and ∠ACF are equilateral. Therefore, they are equiangular (each angle being equal to 60°) and hence ∠BCE ~ ∠ACF.
ar(BCE)/ar(ACF) = BC²/AC²
= BC²/(√2 BC)² [∵ AC = √2 BC]
= BC²/(2BC²) = 1/2
ar(BCE) = (1/2) ar(ACF).
Problem 2:
Question: The areas of two similar triangles ABC and PQR are 64 cm² and 36 cm² respectively. If QR = 16.5 cm, find BC.
Solution:
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of the corresponding sides.
ar(ABC)/ar(PQR) = BC²/QR²
64/36 = BC²/(16.5)²
BC²/272.25 = 16/9
BC² = (16 × 272.25)/9 = 484
BC = √484 = 22
Hence BC = 22 cm.
Problem 3:
Question: In the given figure, LM || BC. AM = 3 cm, MC = 4 cm.
If the ar(ALM) = 27 cm², calculate the ar(ABC).
Solution:
Given: LM || BC AM = 3 cm, MC = 4 cm and ar(ALM) = 27 cm²
To Find: ar(ABC).
Proof:
∠ALM = ∠ABC and ∠AML = ∠ACB [Corresponding ∠s]
∠ALM ~ ∠ABC [By AA criterion of similarity]
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
Therefore,
ar(ALM)/ar(ABC) = AM²/AC²
27/ar(ABC) = 3²/(3 + 4)²
27/ar(ABC) = 9/49
ar(ABC) = (27 × 49)/9
ar(ABC) = 3 × 49
ar (ABC) = 147 cm².
Problem 4:
Question: In the given figure, ABCD is a trapezium in which AB || DC and AB = 2 CD. Find the ratio of the areas of triangles AOB and COD.
Solution:
Given: ABCD is a trapezium in which AB || DC and AB = 2 CD.
To Find: Ratio of the areas of triangles AOB and COD.
Proof:
In ∠AOB and ∠COD,
∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles as AB || DC]
∠AOB ~ ∠COD. [By AA similarity]
Since the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding sides
ar(AOB)/ar(COD) = AB²/CD² [∵ AB = 2 CD]
= (2CD)²/CD² = 4CD²/CD² = 4/1.
Hence, ar(AOB) : ar(COD) = 4 : 1.
PYTHAGORAS THEOREM
Statement:
In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.
Given: A right triangle ABC, right angled at B.
To prove: AC² = AB² + BC².
Construction: Draw BD ⊥ AC.
Proof:
In ∠ADB and ∠ABC, we have
∠ADB = ∠ABC [Each equal to 90°]
∠A = ∠A [Common]
∠ADB ~ ∠ABC [By AA similarity]
AD/AB = AB/AC [Corresponding sides of similar triangles are proportional]
AB² = AC × AD …(i)
In ∠BCD and ∠ACB, we have
∠CDB = ∠CBA [Each equal to 90°]
∠C = ∠C. [Common]
By AA similarity criterion
∠BCD ~ ∠ACB
CD/BC = BC/AC
BC² = AC × CD …(ii)
Adding equations (i) and (ii), we get
AB² + BC² = AC × AD + AC × CD
AB² + BC² = AC(AD + CD)
AB² + BC² = AC × AC
Hence, AC² = AB² + BC². Hence Proved.
CONVERSE OF PYTHAGORAS THEOREM
Statement:
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the first side is a right angle.
Given: A triangle ABC such that AC² = AB² + BC².
Construction: Construct a triangle DEF such that DE = AB, EF = BC and ∠E = 90°.
Proof:
In order to prove that ∠B = 90°, it is sufficient to show ∠ABC ≅ ∠DEF. For this we proceed as follows Since ∠DEF is a right – angled triangle with right angle at E. Therefore, by Pythagoras theorem, we have
DF² = DE² + EF²
DF² = AB² + BC² [∵ DE = AB and EF = BC (By construction)]
DF² = AC² [∵ AB² + BC² = AC² (Given)]
DF = AC .....(i)
Thus, in ∠ABC and ∠DEF, we have
AB = DE, BC = EF. [By construction]
And AC = DF [From equation (i)]
∠ABC ≅ ∠DEF [By SSS criteria of congruency]
∠B = ∠E = 90°.
Hence, ∠ABC is a right triangle, right angled at B.
SOME RESULTS DEDUCED FROM PYTHAGORAS THEOREM:
- In the given figure ABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CD, then AC² = AB² + BC² + 2BC × BD.
2. In the given figure, if ∠B of ∠ABC is an acute angle and AD ⊥ BC, then AC² = AB² + BC² – 2BC × BD.
- In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.
- Three times the sum of the squares of the sides of a triangle is equal to four times the sum of the squares o the medians of the triangle.
SOLVED PROBLEMS
Problem 1:
Question: In a ∠ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that
(i) AD = a√3.
(ii) area (∠ABC) = a²√3.
Solution:
(i) Here, AD ⊥ BC.
Clearly, ∠ABC is an equilateral triangle.
Thus, in ∠ABD and ∠ACD
AD = AD [Common]
∠ADB = ∠ADC. [90° each]
And AB = AC.
by RHS congruency condition
∠ABD ≅ ∠ACD
BD = DC = a.
Now, ∠ABD is a right angled triangle
AD = √(AB² - BD²) [Using Pythagoras Theorem]
AD = √((2a)² - a²)
AD = √(4a² - a²)
AD = √(3a²)
AD = a√3.
(ii) Area (∠ABC) = (1/2) × BC × AD
= (1/2) × 2a × a√3
= a²√3.
Problem 2:
Question: BL and CM are medians of ∠ABC right angled at A. Prove that 4(BL² + CM²) = 5 BC².
Solution:
In ∠BAL
BL² = AL² + AB² ....(i) [Using Pythagoras theorem]
and In ∠CAM
CM² = AM² + AC² .....(ii) [Using Pythagoras theorem]
Adding (i) and (ii) and then multiplying by 4, we get
4(BL² + CM²) = 4(AL² + AB² + AM² + AC²)
= 4{AL² + AM² + (AB² + AC²)} [∵ ∠ABC is a right triangle]
= 4(AL² + AM² + BC²)
= 4(ML² + BC²) [∵ ∠LAM is a right triangle]
= 4ML² + 4 BC²
= 4(BC/2)² + 4BC²
[A line joining mid-points of two sides is parallel to third side and is equal to half of it, ML = BC/2]
= 4(BC²/4) + 4BC²
= BC² + 4BC² = 5BC². Hence proved.
Problem 3:
Question: In the given figure, BC ⊥ AB, AE ⊥ AB and DE ⊥ AC. Prove that DE × BC = AD × AB.
Solution:
In ∠ABC and ∠EDA,
We have
∠ABC = ∠ADE [Each equal to 90°]
∠ACB = ∠EAD [Alternate angles]
By AA Similarity
∠ABC ~ ∠EDA
BC/DE = AB/AD
DE × BC = AD × AB. Hence Proved.
Problem 4:
Question: O is any point inside a rectangle ABCD (shown in the figure). Prove that OB² + OD² = OA² + OC².
Solution:
Through O, draw PQ||BC so that P lies on AB and Q lies on DC.
Now PQ||BC
Therefore PQ ⊥ AB and PQ ⊥ DC [∵ ∠B = 90° and ∠C = 90°]
So, ∠BPQ = 90° and ∠CQP = 90°.
Therefore, BPQC and APQD are both rectangles.
Now, from ∠OPB,
OB² = BP² + OP² ....(i)
Similarly, from ∠ODQ,
OD² = OQ² + DQ² ....(ii)
From ∠OQC, we have
OC² = OQ² + CQ² ...(iii)
And form ∠OAP, we have
OA² = AP² + OP² ....(iv)
Adding (i) and (ii)
OB² + OD² = BP² + OP² + OQ² + DQ²
= CQ² + OP² + OQ² + AP²
[As BP = CQ and DQ = AP]
= CQ² + OQ² + OP² + AP²
= OC² + OA². [From (iii) and (iv)] Hence Proved.
Problem 5:
Question: ABC is a right triangle, right-angled at C. Let BC = a, CA = b, AB = c and let p be the length of perpendicular form C on AB, prove that
(i) cp = ab.
(ii) 1/p² = 1/a² + 1/b².
Solution:
Let CD ⊥ AB. Then CD = p
Area of ∠ABC = (1/2) (Base × height)
= (1/2) (AB × CD) = (1/2)cp.
Also,
area of ∠ABC = (1/2) (BC × AC) = (1/2) ab
(1/2) cp = (1/2) ab
cp = ab.
(ii) Since ∠ABC is a right triangle, right angled at C.
AB² = BC² + AC²
c² = a² + b²
1/c² = 1/(a² + b²)
1/c² = 1/(a² + b²)
(ab)²/c² = (ab)²/(a² + b²)
p² = (ab)²/(a² + b²) [∵ cp = ab ⇒ p = ab/c]
1/p² = (a² + b²)/(ab)²
1/p² = (a² + b²)/(a²b²)
1/p² = a²/(a²b²) + b²/(a²b²)
1/p² = 1/b² + 1/a².
Problem 6:
Question: In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9 AD² = 7AB².
Solution:
ABC be an equilateral triangle and D be point on BC such that
BD = (1/3) BC. (Given)
Draw AE ⊥ BC, Join AD.
BE = EC (Altitude drawn from any vertex of an equilateral triangle bisects the opposite side)
So, BE = EC = BC/2.
In ∠ABC
AB² = AE² + EB² .....(i)
In ∠AED
AD² = AE² + ED² ....(ii)
From (i) and (ii)
AB² = AD² – ED² + EB²
AB² = AD² – (BE - BD)² + EB²
(BD + DE = BE ⇒ DE = BE - BD)
AB² = AD² – (BC/2 - BC/3)² + (BC/2)²
AB² = AD² – (3BC/6 - 2BC/6)² + BC²/4
AB² = AD² – (BC/6)² + BC²/4
AB² = AD² – BC²/36 + BC²/4
AB² = AD² – BC²/36 + 9BC²/36
AB² = AD² + 8BC²/36
AB² = AD² + 2BC²/9
AB² = AD² + 2AB²/9 [∵ BC = AB]
AB² - 2AB²/9 = AD²
(9AB² - 2AB²)/9 = AD²
7AB²/9 = AD²
9AD² = 7AB².
Practice Questions and Solutions
PRACTICE QUESTIONS - SET 1
Questions:
- M and N are points on the sides PQ and PR respectively of ∠PQR. State whether MN || QR. Given PQ = 15.2 cm, PR = 12.8 cm, PM = 5.7 cm, PN = 4.8 cm.
- In the following figure, if AB || DC, find the value of x.
3. In the given figure (i) and (ii) DE || BC. Find EC in (i) and AD in (ii).
4. In the figure, if DE || AC and DF || AE, prove that BF/FE = BE/EC.
5. In the given figure, if DE || OQ and DF || OR, prove that EF || QR.
Answers:
- MN || QR, 2. x = 4, 3. (i) 2 cm, (ii) 2.4 cm
DETAILED SOLUTIONS - SET 1
Solution 1:
Question: M and N are points on the sides PQ and PR respectively of ∠PQR. State whether MN || QR. Given PQ = 15.2 cm, PR = 12.8 cm, PM = 5.7 cm, PN = 4.8 cm.
Solution:
It has been given that
PQ = 15.2 cm, PR = 12.8 cm,
PM = 5.7 cm and PN = 4.8 cm
∴ MQ = PQ – PM = (15.2 – 5.7) cm = 9.5 cm and
NR = PR – PN = (12.8 – 4.8) cm = 8 cm.
Now PM/MQ = 5.7/9.5 = 57/95 = 3/5 and PN/NR = 4.8/8 = 48/80 = 3/5.
∴ PM/MQ = PN/NR.
Thus, in ∠PQR, MN divides the sides PQ and PR in the same ratio. Therefore, by the converse of the Basic Proportionality Theorem, we have MN || QR.
Solution 2:
Question: In the following figure, if AB || DC, find the value of x.
Solution:
Since the diagonals of a trapezium divide each other proportionally
(3x - 1)/(x - 3) = (5x - 3)/(x + 1)
(3x - 1)(x + 1) = (5x - 3)(x - 3)
3x² + 3x - x - 1 = 5x² - 15x - 3x + 9
3x² + 2x - 1 = 5x² - 18x + 9
2x² - 20x + 10 = 0
x² - 10x + 5 = 0
3x(x – 2) + 4 (x – 2) = 0
(x - 2)(3x + 4) = 0
Either x - 2 = 0 or 3x + 4 = 0
x = 2 or x = -4/3.
[x = -4/3 rejected, as it makes line-segments negative]
Thus, x = 4
Solution 3:
Question: In the given figure (i) and (ii) DE || BC. Find EC in (i) and AD in (ii).
Solution:
(i) In ∠ABC, DE || BC then by Basic Proportionality (BPT) Theorem, we have
AD/DB = AE/EC
1.8/5.4 = 1.2/EC
EC = (1.2 × 5.4)/1.8 = 3.6 cm.
(ii) Also in figure (ii), in ∠ABC, DE || BC then by B.P.T, we have
AD/DB = AE/EC
AD/3.6 = 3/4.5
AD = (3 × 3.6)/4.5 = 2.4 cm.
Solution 4:
Question: In the figure, if DE || AC and DF || AE, prove that BF/FE = BE/EC.
Solution:
In ∠ABC, DE || AC then by B.P.T., we have
BD/DA = BE/EC. …(i)
In ∠ABE, DF || AE then by B.P.T., we have
BD/DA = BF/FE. …(ii)
From (i) and (ii), we get
BF/FE = BE/EC.
Solution 5:
Question: In the given figure, if DE || OQ and DF || OR, prove that EF || QR.
Solution:
In ∠POQ, DE || OQ, then by BPT Theorem, we have
PD/DO = PE/EQ …(i)
Also, in ∠POR, DF || OR, then by BPT Theorem, we have
PD/DO = PF/FR. …(ii)
From equations, (i) and (ii),
PE/EQ = PF/FR.
∴ EF || QR [converse of BPT Theorem]
PRACTICE QUESTIONS - SET 2
Questions:
6. In figure, PQ || AB and PR || AC, prove that QR || BC.
7. Using Basic proportionality theorem, prove that the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
8. ABCD is a trapezium such that AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/OD or AO/BO = CO/DO.
9. In figure, prove that AB/AO = CD/OD.
10. Two isosceles triangles have equal vertical angles and their area are in the ratio 16 : 25. Find the ratio of their corresponding heights.
DETAILED SOLUTIONS - SET 2
Solution 6:
Question: In figure, PQ || AB and PR || AC, prove that QR || BC.
Solution:
In ∠POQ, PQ || AB, then by B.P.T., we have
OQ/QA = OP/PB. …(i)
Also in ∠POR, AC || PR, we have
OR/RC = OP/PB. …(ii)
From (i) and (ii), we get
OQ/QA = OR/RC.
∴ QR || BC [by converse of B.P.T.]
Solution 7:
Question: Using Basic proportionality theorem, prove that the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
Given: A ∠ABC in which D is the mid-point of AB and DE || BC meeting AC at E.
To prove: AE = CE.
Proof:
In ∠ABC, DE || BC
AD/BD = AE/CE. [BPT Theorem]
But AD = BD [D is the mid-point of AB]
∴ AE/CE = 1 ⇒ AE = CE.
Hence, E is the midpoint of AC.
Solution 8:
Question: ABCD is a trapezium such that AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/OD or AO/BO = CO/DO.
Solution:
Given: ABCD is a trapezium such that AB || DC.
To prove: AO/OC = BO/OD or AO/BO = CO/DO.
Construction: Through O draw OE || CD.
Proof:
AB || DC [given]
and EO || DC [const.]
∴ EO || AB.
(∵ Two lines parallel to same line are parallel to each other)
Now, in ∠ABD, EO || AB, then by BPT Theorem we have
DO/OB = DE/EA …(i)
Also in ∠ADC, EO || DC, by BPT Theorem, we have
AO/OC = AE/ED …(ii)
From (i) and (ii), we get
AO/OC = BO/OD or AO/BO = CO/DO.
Solution 9:
Question: In figure, prove that AB/AO = CD/OD.
Solution:
Construction: Draw AX and DY ⊥ BC.
Proof:
In ∠AOX and ∠DOY, we have
∠AXO = ∠DYO [Both 90°]
∠AOX = ∠DOY [Vert. Opp. angles]
∠AOX ~ ∠DOY [By AA similarity]
∴ AO/OD = AX/DY …(i)
Now ar(∠AOB)/ar(∠DOC) = (1/2 × AB × AX)/(1/2 × CD × DY) = (AB × AX)/(CD × DY).
Thus ar(∠AOB)/ar(∠DOC) = (AB/CD) × (AX/DY) = (AB/CD) × (AO/OD). [Using (i)]
Solution 10:
Question: Two isosceles triangles have equal vertical angles and their area are in the ratio 16 : 25. Find the ratio of their corresponding heights.
Solution:
Given: Let ∠ABC and ∠DEF be the given triangles such that AB = AC and DE = DF, ∠A = ∠D
and, ar(∠ABC)/ar(∠DEF) = 16/25 …(i)
To Find: AL/DM
Construction: Draw AL ⊥ BC and DM ⊥ EF.
Proof:
Now, AB = AC, DE = DF
∴ AB/AC = 1 and DE/DF = 1
∴ AB/AC = DE/DF
Thus, in triangles ABC and DEF, we have
AB/DE = AC/DF and ∠A = ∠D. [Given]
So by SAS-similarity criterion, we have
∠ABC ~ ∠DEF
∴ ar(∠ABC)/ar(∠DEF) = AL²/DM²
[Ratio of areas of two similar triangles is equal to ratio of squares of their corresponding altitudes]
16/25 = AL²/DM² [Using (i)]
∴ AL/DM = 4/5.
PRACTICE QUESTIONS - SET 3
11. A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.
12. A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.
13. In the given figure, ∠ACB = 90° and CD ⊥ AB. Prove that BC²/AC² = BD/AD.
14. ABC is a triangle right-angled at C and p is the length of the perpendicular from C to AB. Show that (a) pc = ab. (b) 1/p² = 1/a² + 1/b² where a = BC, b = AC and c = AB.
15. In ∠ABC, ∠C > 90° and side AC is produced to D such that segment BD is perpendicular to segment AD. Prove that AB² = AC² + BC² + 2AC.CD
DETAILED SOLUTIONS - SET 3
Question: A ladder is placed in such a way that its foot is at a distance of 5 m from a wall and its top reaches a window 12 m above the ground. Determine the length of the ladder.
Solution:
Let AB be the ladder, B be the window and BC be the wall.
Then BC = 12 m, AC = 5m and ∠ACB = 90°.
In right triangle ACB, we have
AB² = BC² + AC² [By Pythagoras Theorem]
AB² = (12)² + (5)²
AB² = 144 + 25 = 169
AB = √169 = 13 m.
Hence the length of the ladder is 13 m.
Question: A ladder 15 m long reaches a window which is 9 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.
Solution:
Let AB be the street and let C and D be the windows at heights of 9 m and 12 m respectively from the ground.
Let E be the foot of the ladder. Then EC and ED are the two positions of the ladder.
Clearly AC = 9 m, BD = 12 m, EC = ED = 15 m and ∠CAE = ∠DBE = 90°
In right triangle CAE, we have
EC² = AC² + AE² [By Pythagoras Theorem]
(15)² = (9)² + AE²
AE² = (15)² – (9)² = (225 – 81) = 144
AE = 12 m. …(i)
In right triangle DBE, we have
ED² = BD² + EB² [By Pythagoras Theorem]
(15)² = (12)² + EB²
EB² = (15)² – (12)² = (225 – 144) = 81
EB = 9 m …(ii)
Adding equations (i) and (ii), we get
AE + EB = (12 + 9) m ⇒ AB = 21 m
Hence, the width of the street is 21 m.
Question: In the given figure, ∠ACB = 90° and CD ⊥ AB. Prove that BC²/AC² = BD/AD.
Solution:
Given: ∠ACB = 90° and CD ⊥ AB.
To prove: BC²/AC² = BD/AD.
Proof:
In ∠ACD and ∠ABC
∠A = ∠A [Common]
∠ADC = ∠ACB [Both 90°]
∠ACD ~ ∠ABC [By AA similarity]
So AC/AB = AD/AC
or, AC² = AB × AD. …(i)
Similarly, ∠BCD ~ ∠BAC.
So BC/AB = BD/BC
or, BC² = AB × BD. …(ii)
Therefore, from (i) and (ii),
BC²/AC² = (AB × BD)/(AB × AD) = BD/AD.
Question: ABC is a triangle right-angled at C and p is the length of the perpendicular from C to AB. Show that (a) pc = ab. (b) 1/p² = 1/a² + 1/b² where a = BC, b = AC and c = AB.
Solution:
(a) Taking c as the base and p as the altitude, we have
area of ∠ABC = (1/2) × c × p …(i)
Taking b as the base and a as the altitude, we have
area ∠ABC = (1/2) × a × b …(ii)
∴ (1/2) × c × p = (1/2) × a × b [From (i) and (ii)]
⇒ pc = ab Hence Proved.
(b) ∠ABC is a right triangle-angled at C.
∴ c² = a² + b² …(iii)
[By Pythagoras Theorem]
pc = ab [proved above]
or p = ab/c
∴ p² = (ab)²/c² [From equation (iii)]
p² = (ab)²/(a² + b²)
1/p² = (a² + b²)/(ab)²
1/p² = (a² + b²)/(a²b²)
1/p² = a²/(a²b²) + b²/(a²b²)
1/p² = 1/b² + 1/a².
Frequently Asked Questions
Congruent triangles are triangles that are identical in both shape and size. This means all three corresponding sides have equal lengths and all three corresponding angles have equal measures. When triangles are congruent, you can essentially place one triangle on top of the other and they will match perfectly. The symbol for congruence is ≅, so we write △ABC ≅ △DEF. Congruence can be proven using several criteria: SSS (all three sides equal), SAS (two sides and the included angle equal), ASA (two angles and the included side equal), AAS (two angles and a non-included side equal), or RHS (right angle, hypotenuse, and one side equal for right triangles).
Similar triangles, on the other hand, have the same shape but not necessarily the same size. They are essentially scaled versions of each other. For triangles to be similar, two conditions must be met: all corresponding angles must be equal, and all corresponding sides must be proportional (in the same ratio). The symbol for similarity is ~, written as △ABC ~ △DEF. Similar triangles maintain their shape when enlarged or reduced, which is why blueprints, maps, and scale models rely on this principle. The key difference is that congruent triangles have a scale factor of 1:1, while similar triangles can have any scale factor such as 2:1, 3:2, or 5:3.
Understanding this distinction is crucial because congruent triangles have equal areas and equal perimeters, while similar triangles have areas in the ratio of the square of their corresponding sides and perimeters in the ratio of their corresponding sides. In practical applications, similarity is used more frequently because it allows us to work with proportional relationships, such as calculating the height of a building using shadows or creating architectural scale drawings.
The Basic Proportionality Theorem, also known as Thales' Theorem, states that if a line is drawn parallel to one side of a triangle and intersects the other two sides at distinct points, then it divides those two sides proportionally. Mathematically, in triangle ABC, if a line DE is drawn such that D is on side AB, E is on side AC, and DE is parallel to BC, then AD/DB = AE/EC. This theorem is foundational in geometry because it establishes the relationship between parallel lines and proportional segments.
You should use this theorem whenever you encounter a triangle with a line segment parallel to one of its sides. The most common applications include: finding unknown lengths in triangles when parallel lines are present, proving that certain lines are parallel (using the converse of the theorem), solving problems involving midpoints (since the line joining midpoints of two sides is parallel to the third side), and working with trapeziums where parallel sides create proportional relationships.
The corollary of BPT is equally useful: AD/DB = AE/EC can be rewritten as AB/DB = AC/EC by adding 1 to both sides. This alternative form often simplifies calculations because it relates the entire side to one of its segments rather than comparing two segments. The converse of BPT states that if a line divides two sides of a triangle in the same ratio, then that line must be parallel to the third side. This is particularly valuable for proving parallel relationships.
In practical problem-solving, look for these indicators that BPT applies: you see a triangle with a line segment cutting through two sides, you're given ratios of segments, you need to prove lines are parallel, or you're working with similar triangles (which are closely related to BPT). The theorem also extends to multiple parallel lines intersected by transversals, where the intercepts on the transversals are proportional. Mastering BPT enables you to solve complex geometric problems involving proportionality, similarity, and parallel line relationships with confidence and efficiency.
There are three primary criteria for establishing triangle similarity, each offering a different approach depending on the information available. Understanding when to apply each criterion is essential for efficient problem-solving in geometry.
AAA (Angle-Angle-Angle) Similarity Criterion: Two triangles are similar if all three corresponding angles are equal. However, since the sum of angles in any triangle equals 180°, you only need to prove that two pairs of corresponding angles are equal—the third pair automatically becomes equal. This criterion is particularly useful when working with parallel lines (which create equal corresponding or alternate angles), when triangles share a common angle, or when angle relationships are more readily available than side measurements. In real-world applications, this criterion is used in surveying, navigation, and situations where angles are easier to measure than distances.
SSS (Side-Side-Side) Similarity Criterion: Two triangles are similar if their corresponding sides are proportional. This means AB/DE = BC/EF = CA/FD for triangles ABC and DEF. This criterion is most appropriate when you have information about all six sides of the two triangles, when working with scale drawings or models, or when measurements are given without angle information. The advantage of this criterion is that it's purely computational—you calculate the ratios and check if they're equal. It's extensively used in architecture, engineering, and design where proportional scaling is essential.
SAS (Side-Angle-Side) Similarity Criterion: Two triangles are similar if one pair of corresponding angles is equal and the sides including these angles are proportional. This criterion provides a middle ground between AAA and SSS, requiring less information than SSS but more than AAA. It's particularly efficient when you have mixed information—some angles and some sides—which is common in practical problems. This criterion is often used in problems involving angle bisectors, where the angle condition is easily established and you need to verify the proportionality of the adjacent sides.
When approaching a similarity problem, first identify what information is given. If you have all angles or can easily determine them, use AAA. If you have measurements of all sides, use SSS. If you have one angle and adjacent side information, SAS is most efficient. Remember that proving similarity unlocks powerful relationships: corresponding sides become proportional, areas relate by the square of the side ratio, and corresponding altitudes, medians, and angle bisectors are also proportional. These relationships make similarity one of the most versatile tools in geometric problem-solving.
The Pythagorean Theorem is one of the most fundamental relationships in mathematics, stating that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) equals the sum of the squares of the other two sides. Mathematically, if triangle ABC has a right angle at C, then AB² = AC² + BC², where AB is the hypotenuse.
Application Process: First, verify you have a right triangle or create one by drawing an altitude. Identify the hypotenuse (always the longest side, opposite the right angle). Label the other two sides as legs or catheti. Substitute the known values into the formula c² = a² + b². Solve for the unknown side by taking the square root if solving for a side length, or by squaring and adding if verifying the relationship. Always check that your answer is reasonable—the hypotenuse must be longer than either leg.
Common Applications include: Distance calculations in coordinate geometry, where the distance formula is derived directly from the Pythagorean Theorem. For two points (x₁, y₁) and (x₂, y₂), the distance equals √[(x₂ − x₁)² + (y₂ − y₁)²]. Navigation problems, where vehicles travel in perpendicular directions (like north then east) and you need to find the direct distance. Construction and carpentry, where the theorem ensures right angles using the 3-4-5 rule or its multiples (if sides measure 3, 4, and 5 units, the triangle is right-angled). Height and distance problems, such as finding the height of a ladder against a wall or the diagonal of a rectangle. Three-dimensional geometry, where the theorem extends to find space diagonals in rectangular boxes.
The converse of the Pythagorean Theorem is equally important: if the square of one side of a triangle equals the sum of the squares of the other two sides, then the triangle is right-angled at the vertex opposite that side. This allows you to verify whether a triangle is right-angled when only side lengths are known. For example, a triangle with sides 5, 12, and 13 is right-angled because 13² = 5² + 12² (169 = 25 + 144).
Extended relationships derived from the Pythagorean Theorem include formulas for obtuse and acute triangles. For an obtuse triangle ABC with obtuse angle at B and altitude AD drawn to BC extended: AC² = AB² + BC² + 2BC·BD. For an acute angle: AC² = AB² + BC² − 2BC·BD. Understanding these extensions allows you to apply Pythagorean relationships to all triangles, not just right triangles, making it one of the most versatile tools in your mathematical toolkit.
This relationship is one of the most powerful theorems in geometry, connecting linear measurements (sides) to two-dimensional measurements (areas). Understanding why this relationship holds requires exploring the fundamental nature of area scaling.
Conceptual Understanding: When you scale a triangle by a factor k (meaning each side is multiplied by k), the resulting triangle is similar to the original. However, area is a two-dimensional measurement, so it scales by k². Think of it this way: if you double the sides of a triangle (k = 2), you don't double the area—you quadruple it (2² = 4). If you triple the sides (k = 3), the area multiplies by 9 (3² = 9). This is because area involves multiplication of two dimensions (like base times height), and each dimension scales by k.
Mathematical Proof: Consider two similar triangles ABC and PQR with similarity ratio AB/PQ = BC/QR = CA/RP = k. Draw altitudes AM and PN from vertices A and P to the opposite sides. Since the triangles are similar, these altitudes are also in the ratio k (corresponding altitudes of similar triangles are proportional). The area of triangle ABC = (1/2) × BC × AM, and the area of triangle PQR = (1/2) × QR × PN. Taking the ratio: Area(ABC)/Area(PQR) = (BC × AM)/(QR × PN) = (BC/QR) × (AM/PN) = k × k = k². This proves that the area ratio equals the square of the side ratio.
Practical Implications: This theorem has numerous applications. In scale models and drawings, if an architectural drawing is made at 1:50 scale (linear), the area of rooms on the drawing is 1:2500 of actual area (1²:50² = 1:2500). In similar triangle problems, once you establish similarity, you can immediately write the area relationship without recalculating areas from scratch. In optimization problems, understanding how area scales helps determine efficient designs. For map reading, if a map scale is 1 cm = 10 km (linear), then 1 cm² on the map represents 100 km² in reality (scale² = 10² = 100).
Extended Properties: This quadratic scaling relationship extends beyond just sides to areas. The ratio of areas also equals the square of the ratio of corresponding medians, altitudes, or angle bisector segments. If two similar triangles have medians in ratio 2:3, their areas are in ratio 4:9. This provides multiple pathways to solving problems—you can work with whatever corresponding linear measurements are most convenient, knowing the area ratio will always be the square of that linear ratio.
Understanding this relationship is crucial for advanced geometry, trigonometry, and calculus, where area scaling principles underpin integration, surface area calculations, and dimensional analysis. It also connects to the broader mathematical principle that n-dimensional measurements scale by the nth power of the linear scale factor—lengths scale by k, areas by k², and volumes by k³.
The Angle Bisector Theorem is a powerful result about the proportional relationships created when an angle bisector is drawn in a triangle. It states that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the two sides containing the angle. In triangle ABC, if AD is the bisector of angle A (meeting BC at D), then BD/DC = AB/AC.
Understanding the Theorem: This theorem reveals that the angle bisector doesn't simply cut the opposite side into two equal parts (which would only happen in an isosceles triangle). Instead, it creates a division proportional to the adjacent sides. If AB = 10 cm and AC = 6 cm, and the angle bisector from A meets BC at D, then BD/DC = 10/6 = 5/3. This means if DC = 3 units, then BD = 5 units, making BC = 8 units total. The side closer to the longer adjacent side (AB) gets the larger segment.
Proof Approach: The theorem can be proven using similar triangles. Draw a line through C parallel to AD, extending AB to meet this line at E. Then triangle ADE has AE = AC (since triangle ADC is isosceles due to the parallel lines), and by the Basic Proportionality Theorem, BD/DC = AB/AE = AB/AC. This elegant proof connects the Angle Bisector Theorem to proportionality and similarity principles.
Practical Applications include: Finding unknown lengths when an angle bisector is present and two sides are known. Proving that a line is an angle bisector by showing it divides the opposite side in the ratio of the adjacent sides (the converse). Construction problems where you need to divide a side in a specific ratio this can be accomplished geometrically using an angle bisector. Optimization problems in which angle bisectors play a role in minimizing distances or creating specific proportional relationships.
The External Angle Bisector Theorem is the complement: the external bisector of an angle divides the opposite side externally in the ratio of the sides containing the angle. If the external bisector of angle A meets BC extended at point D', then BD'/D'C = AB/AC. Note that D' lies outside the segment BC. This external version is useful in problems involving excircles and external division points.
Connection to Other Theorems: The Angle Bisector Theorem is closely related to similarity and the Basic Proportionality Theorem. Many problems combine these concepts for instance, showing that if two triangles have proportional sides and one has an angle bisector dividing the opposite side in the same ratio as the other triangle's corresponding bisector, then specific similarity relationships hold. Understanding these interconnections allows you to approach complex geometry problems with multiple tools.
When approaching problems, look for these indicators: an angle bisector is mentioned or drawn, you need to find the ratio of segments on a side, you know two sides of a triangle and need information about the third, or you need to prove a line bisects an angle. The theorem provides a direct computational method that often bypasses more complex similarity or trigonometric approaches, making it an essential tool for efficient problem-solving in triangle geometry.
Determining whether a triangle is right-angled without directly measuring angles is accomplished using the converse of the Pythagorean Theorem. This states: if the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is right-angled, with the right angle opposite the longest side.
Step-by-Step Process: First, identify the three side lengths and arrange them in order: a ≤ b ≤ c, where c is the longest side. Calculate the squares: compute a², b², and c². Check the relationship: calculate a² + b² and compare it to c². If a² + b² = c², the triangle is right-angled with the right angle opposite side c. If a² + b² > c², the triangle is acute-angled (all angles less than 90°). If a² + b² < c², the triangle is obtuse-angled (one angle greater than 90°).
Common Pythagorean Triples: Certain sets of three integers satisfy the Pythagorean relationship and are worth memorizing: (3, 4, 5) is the most fundamental triple. Since 3² + 4² = 9 + 16 = 25 = 5², any triangle with these side lengths is right-angled. All multiples are also Pythagorean triples: (6, 8, 10), (9, 12, 15), (15, 20, 25), etc. (5, 12, 13) is another primitive triple, along with its multiples. (8, 15, 17), (7, 24, 25), and (20, 21, 29) are additional useful triples. Recognizing these patterns allows you to quickly identify right triangles without calculation.
Practical Examples: Consider a triangle with sides 7, 24, and 25. Checking: 7² + 24² = 49 + 576 = 625 = 25². Therefore, this is a right triangle with the right angle opposite the side of length 25. For a triangle with sides 5, 6, and 8: 5² + 6² = 25 + 36 = 61, while 8² = 64. Since 61 < 64, this is an obtuse triangle. For sides 6, 7, and 8: 6² + 7² = 36 + 49 = 85, while 8² = 64. Since 85 > 64, this is an acute triangle.
Real-World Applications: In construction, the 3-4-5 method is used to establish right angles: measure 3 units along one direction, 4 units perpendicular to that, and if the diagonal measures exactly 5 units, you have a perfect right angle. In surveying and navigation, determining whether terrain forms right angles helps in land division and property boundaries. In quality control manufacturing, verifying that corners are truly 90° can be done by measuring the three sides of the triangular space and applying this test. In carpentry and woodworking, ensuring squares and frames are true right angles prevents fitting problems later in construction.
Extending the Concept: The Pythagorean relationship also helps calculate the third side when two sides are known. If you have a right triangle with legs of 9 cm and 12 cm, the hypotenuse c satisfies c² = 9² + 12² = 81 + 144 = 225, so c = 15 cm. If you know the hypotenuse is 20 cm and one leg is 12 cm, the other leg b satisfies b² = 20² − 12² = 400 − 144 = 256, so b = 16 cm. This calculation method is essential for indirect measurement problems where direct measurement is impractical.
Solving complex triangle problems requires systematic thinking and the ability to recognize which theorems apply. Developing a strategic approach transforms seemingly difficult problems into manageable steps.
Initial Analysis: Begin by carefully reading the problem and identifying what is given and what needs to be found. Draw an accurate diagram, labeling all given information clearly. Mark equal angles, equal sides, right angles, and parallel lines using standard notation. Look for special triangle types: right triangles (enabling Pythagorean Theorem), isosceles triangles (equal sides and angles), equilateral triangles (all sides and angles equal), or similar triangles (proportional relationships).
Identify Key Relationships: Check for parallel lines, which create equal corresponding angles, equal alternate angles, and trigger the Basic Proportionality Theorem. Look for angle bisectors, which create proportional divisions via the Angle Bisector Theorem. Identify medians (connecting vertices to midpoint of opposite sides), which can be analyzed using the Midpoint Theorem or median length formulas. Recognize altitudes (perpendicular to sides), which create right triangles suitable for Pythagorean applications. Search for similar triangles by checking if corresponding angles are equal or sides are proportional.
Strategic Theorem Selection: When parallel lines and proportionality are involved, apply the Basic Proportionality Theorem or its corollary. When you need to find a third side and have a right triangle, use the Pythagorean Theorem. When working with similar triangles, establish similarity first (using AAA, SSS, or SAS criteria), then use proportional sides and area relationships. When angle bisectors appear, apply the Angle Bisector Theorem to establish side ratios. When areas are involved, remember that areas of similar triangles relate by the square of the side ratio. When multiple approaches seem possible, choose the path with the fewest unknowns and most direct relationships.
Common Problem Patterns:
Shadow problems: Objects and their shadows create similar triangles when measured simultaneously. Set up proportions using object heights and shadow lengths.
Ladder problems: A ladder against a wall forms a right triangle; use Pythagorean Theorem with the wall height, ground distance, and ladder length.
Navigation problems: Perpendicular movements create right triangles; find the direct distance using the Pythagorean Theorem.
Trapezium problems: Parallel sides and intersecting diagonals create proportional segments and similar triangles.
Midpoint problems: Lines joining midpoints are parallel to the third side and half its length; use Midpoint Theorem.
Multi-Step Problem Approach: Break complex problems into smaller sub-problems. Solve for intermediate unknowns that lead to the final answer. Use auxiliary lines (additional lines drawn to create useful triangles or relationships)—common choices include drawing altitudes to create right triangles, drawing lines parallel to given sides to create similar triangles, or connecting midpoints to use the Midpoint Theorem. Verify answers using alternative methods when possible, checking that calculated values satisfy all given conditions and make physical sense.
Common Mistakes to Avoid: Don't assume angles are right angles without verification check for the right angle symbol or verify using the Pythagorean converse. Don't confuse similarity with congruence similar triangles have proportional sides, not necessarily equal sides. Remember to check that you're using corresponding sides in similarity proportions match angles to sides correctly. When using square roots, remember that side lengths must be positive. In problems involving ratios, keep track of whether you're working with segment ratios or whole side ratios.
Advanced Integration: In sophisticated problems, multiple theorems work together. You might use BPT to establish proportional segments, then use those proportions to prove triangle similarity, then use the area ratio theorem to find an unknown area. Or you might apply the Angle Bisector Theorem to find a side ratio, use that to establish similarity with another triangle, then apply Pythagorean Theorem to a resulting right triangle. Building fluency with these theorem combinations comes through deliberate practice with diverse problem types.
Developing strong problem-solving strategies in triangle geometry builds mathematical reasoning skills that extend beyond geometry into algebra, calculus, physics, and engineering. The key is systematic analysis, pattern recognition, and strategic theorem application skills that improve with intentional practice and reflection on both successful and unsuccessful problem-solving attempts.