Triangles

(A) Scalene Triangle:

A triangle whose no two sides are equal is called a scalene triangle.

(B) Isosceles Triangle:

A triangle having two sides equal is called an isosceles triangle.

(C) Equilateral Triangle:

A triangle in which all sides are equal is called an equilateral triangle.

(A) Right Triangle:

A triangle in which any one angle is right angle (=90°) is called right triangle.

(B) Acute Triangle:

A triangle in which all angles are acute (>90°) is called an acute triangle.

(C) Obtuse Triangle:

A triangle in which any one angle is obtuse (<90°) is called an obtuse triangle.

The figures are called congruent if they have same shape and same size. In other words, two figures are called congruent if they are having equal length, width and height.

Two triangles are congruent if and only if one of them can be made to superimposed on the other, so to cover it exactly.

If two triangles ABC and DEF are congruent then there exist a one to one correspondence between their vertices and sides. i.e. we get following six equalities:

∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AB = DE, BC = EF, AC = DF

Notation: If two ΔABC & ΔDEF are congruent under A ↔ D, B ↔ E, C ↔ F one to one correspondence then we write ΔABC ≅ ΔDEF

We cannot write as ΔABC ≅ ΔDFE of ΔABC ≅ ΔEDF or in other forms because ΔABC ≅ ΔDFE have following one-one correspondence A ↔ D, B ↔ F, C ↔ E.

Two triangles are congruent if and only if there exists a one-one correspondence between their vertices such that the corresponding sides and the corresponding angles of the two triangles are equal.

SAS Congruence Criterion (Side-Angle-Side):

Statement: Two triangles are congruent if two sides and the included angle of one are equal to the corresponding sides and the included angle of the other triangle.

Example 1:

Given: O is the mid-point of AB and CD.

To Prove:

1. ΔAOC ≅ ΔBOD
2. AC = BD
3. AC || BD

Solution: In Δs AOC and BOD, we have

• AO = OB [∵ O is the mid-point of AB]
• ∠AOC = ∠BOD [Vertically opposite angles]
• CO = OD [∵ O is the mid-point of CD]

So, by SAS congruence criterion, we have ΔAOC ≅ ΔBOD

⇒ AC = BD and ∠CAO = ∠DBO [Corresponding parts of congruent triangles are equal]

Now, AC and BD are two lines intersected by a transversal AB such that ∠CAO = ∠DBO, i.e. alternate angles are equal. Therefore, AC || BD

ASA Congruence Criterion (Angle-Side-Angle):

Statement: Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

Given: Two Δs ABC and DEF such that ∠B = ∠E, ∠C = ∠F and BC = EF

To prove: ΔABC ≅ ΔDEF

Proof: There are three possibilities

Case I: When AB = DE:

In this case, we have AB = DE, ∠B = ∠E, and BC = EF

So, by SAS criterion of congruence, we have ΔABC ≅ ΔDEF

Case II: When BA < ED

In this case take a point G on ED such that EG = BA. Join GF.

Now, in Δs ABC and GEF, we have:

• BA = GE [Assumed]
• ∠B = ∠E [Given]
• BC = EF [Given]

So, by SAS criterion of congruence, we have ΔABC ≅ ΔGEF

⇒ ∠ACB = ∠GFE [cpct]

But, ∠ACB = ∠DFE [Given]

∴ ∠GFE = ∠DFE

This is possible only when ray FG coincides with ray FD or G coincides with D. Therefore, BA must be equal to ED.

Thus, in Δs ABC and DEF, we have AB = DE [As proved above], ∠B = ∠E [Given], and BC = EF [Given]

So, by SAS criterion of congruence, we have ΔABC ≅ ΔDEF

Case III: when BA > ED

In this case take a point G on ED produced such that EG = BA. Join GF. Now, proceeding as in case II, we prove that ΔABC ≅ ΔDEF

Hence, in all the three cases, we have ΔABC ≅ ΔDEF

AAS Congruence Criterion (Angle-Angle-Side):

Statement: If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

Given: Two Δs ABC and DEF such that ∠A = ∠D, ∠B = ∠E, BC = EF

To prove: ΔABC ≅ ΔDEF

Proof: We have, ∠A = ∠D, and ∠B = ∠E

• ⇒ ∠A + ∠B = ∠D + ∠E
• ⇒ 180° − ∠C = 180° − ∠F [Angle sum property]
• ⇒ ∠C = ∠F ...(i)

Thus, in Δs ABC and DEF, we have: ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Now, in ΔABC and ΔDEF, we have:

• ∠B = ∠E [Given]
• BC = EF [Given]
• ∠C = ∠F [From (i)]

So, by ASA criterion of congruence, ΔABC ≅ ΔDEF

SSS Congruence Criterion (Side-Side-Side):

Statement: Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

Given: Two Δs ABC and DEF such that AB = DE, BC = EF and AC = DF.

To prove: ΔABC ≅ ΔDEF

Construction: Suppose BC is the longest side. Draw EG such that ∠FEG = ∠ABC and EG = AB. Join GF and GD.

Proof: In Δs ABC and ΔGEF, we have:

• BC = EF [Given]
• AB = GE [By construction]
• ∠ABC = ∠FEG [By construction]

So, by SAS criterion of congruence, we have ΔABC ≅ ΔGEF

⇒ ∠A = ∠G and AC = GF [c.p.c.t.]

Now, AB = DE and AB = GE ⇒ DE = GE

Similarly, AC = DF and AC = GF ⇒ DF = GF

In ΔEGD, we have: DE = GE [From above]

⇒ ∠EDG = ∠EGD [Angles opp. to equal side]

In ΔFGD, we have: DF = GF [From above]

⇒ ∠FDG = ∠FGD [Angles opp. to equal side]

Adding the above two equations:

∠EDG + ∠FDG = ∠EGD + ∠FGD

∴ ∠D = ∠G

But, ∠G = ∠A [Proved above] ∴ ∠A = ∠D

Thus, in Δs ABC and DEF, we have:

• AB = DE [Given]
• ∠A = ∠D [From above]
• AC = DF [Given]

So, by SAS criterion of congruence, we have ΔABC ≅ ΔDEF

RHS Congruence Criterion (Right Angle-Hypotenuse-Side):

Statement: Two right triangles are congruent if the hypotenuse and one side of one triangle are respectively equal to the hypotenuse and one side of the other triangle.

Given: Two right angles ABC and DEF in which ∠B = ∠E = 90°, AC = DF, BC = EF

To prove: ΔABC ≅ ΔDEF

Construction: Produce DE to G so that EG = AB. Join GF

Proof: In Δs ABC and ΔGEF, we have:

• AB = GE [By construction]
• ∠ABC = ∠FEG [Both 90°]
• BC = EF [Given]

So, by SAS criterion of congruence, we have ΔABC ≅ ΔGEF

⇒ ∠A = ∠G ...(i) and AC = GF [cpct] ...(ii)

Now, AC = GF [From (ii)] and AC = DF [Given] ∴ DF = GF

⇒ ∠D = ∠G [Angles opposite to equal sides] ...(iii)

From (i) and (iii), we get ∠A = ∠D ...(iv)

Thus, in Δs ABC and DEF, we have:

• ∠A = ∠D [From (iv)]
• ∠B = ∠E [Both 90°]
• ⇒ ∠A + ∠B = ∠D + ∠E
• ⇒ 180° − ∠C = 180° − ∠F [Angle sum property]
• ⇒ ∠C = ∠F ...(v)

Now, in Δs ABC and DEF, we have:

• BC = EF [Given]
• ∠C = ∠F [From (v)]
• AC = DF

So, by SAS criterion of congruence, we have ΔABC ≅ ΔDEF

THEOREM 1:

Statement: Angles opposite to equal sides of an isosceles triangle are equal.

Given: An isosceles triangle ABC in which AB = AC.

To prove: ∠B = ∠C.

Construction: Draw the bisector of ∠A and let D be the point of intersection of this bisector of ∠A and BC.

Proof: In ΔBAD and ΔCAD

• AB = AC [Given]
• ∠BAD = ∠CAD [By construction]
• AD = AD [Common]

ΔBAD ≅ ΔCAD [By SAS Congruency]

∴ ∠ABD = ∠ACD [cpct]

∴ ∠B = ∠C

THEOREM 2:

Statement: If two angles of a triangle are equal, then sides opposite to them are also equal.

Given: Δ ABC in which ∠B = ∠C

To prove: AC = AB

Construction: Draw the bisector of ∠A and let it meet BC at D.

Proof: In Δs ABD and ACD, we have:

• ∠B = ∠C [Given]
• ∠BAD = ∠CAD [By construction]
• AD = AD [Common]

So, by SAS criterion of congruence, we have ΔABD ≅ ΔACD

⇒ AB = AC [cpct]

Hence, ΔABC is isosceles.

THEOREM 3:

Statement: The sum of any two sides of a triangle is greater than the third side.

Given: Δ ABC

To prove: AB + AC > BC, AB + BC > AC and BC + AC > AB

Construction: Extend side BA to D such that AD = AC. Join CD

Proof: In ΔACD, we have:

• AC = AD [By construction]
• ⇒ ∠ADC = ∠ACD [Angles opp. to equal sides are equal]
• ⇒ ∠ACD = ∠ADC

∴ ∠BCA + ∠ACD > ∠ADC [∵ ∠BCA + ∠ACD > ∠ACD]

⇒ ∠BCD > ∠ADC

⇒ ∠BCD > ∠BDC [∵ ∠ADC = ∠BDC]

⇒ BD > BC [∵ Side opp. to greater angle is larger]

⇒ BA + AD > BC

⇒ BA + AC > BC [∵ AC = AD (By construction)]

⇒ AB + AC > BC

Thus, AB + AC > BC

Similarly, AB + BC > AC and BC + AC > AB

Inequality Relation 1:

Statement: If two sides of a triangle are unequal, the longer side has greater angle opposite to it.

Given: Δ ABC in which AC > AB

To prove: ∠ABC > ∠ACB

Construction: Mark a point D on AC such that AB = AD. Join BD.

Proof: In ΔABD, we have:

• AB = AD [By construction]
• ⇒ ∠1 = ∠2 [∵ Angles opp. to equal sides are equal] ...(i)

∠2 is the exterior angle of ΔBCD and an exterior angle is always greater than interior opposite angle. Therefore,

∠2 > ∠DCB

⇒ ∠2 > ∠ACB [∵ ∠ACB = ∠DCB] ...(ii)

From (i) and (ii), we have:

⇒ ∠1 > ∠ACB ...(iii)

∵ ∠ABC > ∠1 ...(iv)

From (iii) and (iv), we get ∠ABC > ∠ACB

Inequality Relation 2:

Statement: In a triangle the greater angle has the longer side opposite to it.

Given: Δ ABC in which ∠ABC > ∠ACB

To prove: AC > AB

Proof: In ΔABC, we have the following three possibilities:

1. AC = AB
2. AC < AB
3. AC > AB

Case I: When AC = AB

If AC = AB ⇒ ∠ABC = ∠ACB [Angle opp. to equal sides are equal]

This is a contradiction, Since ∠ABC > ∠ACB [Given]

∴ AC ≠ AB

Case II: When AC < AB

if AC < AB ⇒ ∠ACB > ∠ABC [∵ Longer sides has the greater angle opposite to it]

This is also a contradiction

∴ We are left with the only possibility, AC > AB, which must be true

Hence, AC > AB

Example 1:

Question: Prove that ΔABC is isosceles if any one of the following holds:

1. Altitude AD bisects BC
2. Median AD is perpendicular to the base BC

Solution (i):

Let ΔABC be a triangle and let AD, be altitude from A to BC. Suppose D bisects BC i.e. BD = CD. We have to prove that the ΔABC is isosceles.

In ΔADB and ΔADC, we have:

• AD = AD [Common side]
• ∠ADB = ∠ADC [Each equal to 90°]
• BD = DC [Given]

So, by SAS criterion of congruence, we have ΔADB ≅ ΔADC

⇒ AB = AC [cpct]

Hence, ΔABC is isosceles.

Solution (ii):

Let ΔABC be a triangle such that the median AD perpendicular to the base BC.

In Δs ADB and ADC, we have:

• AD = AD [Common side]
• ∠ADB = ∠ADC [Both 90°]
• BD = DC [∵ AD is the median ∴ D is the mid-point of BC]

So, by SAS criterion of congruence, we have ΔADB ≅ ΔADC

⇒ AB = AC [cpct]

Hence, ΔABC is isosceles.

Example 2:

Question: In figure, PQRS is a quadrilateral and T and U are respectively points on PS and RS such that PQ = RQ, ∠PQT = ∠RQU and ∠TQS = ∠UQS. Prove that QT = QU.

Solution:

We have, ∠PQT = ∠RQU and ∠TQS = ∠UQS

⇒ ∠PQT + ∠TQS = ∠RQU + ∠UQS

⇒ ∠PQS = ∠RQS ...(i)

Thus, in triangles PQS and RQS, we have:

• PQ = RQ [Given]
• ∠PQS = ∠RQS [From (i)]
• QS = QS [Common side]

Therefore, by SAS congruence criterion, we have ΔPQS ≅ ΔRQS

⇒ ∠QPS = ∠QRS [cpct]

⇒ ∠QPT = ∠QRU ...(ii)

In Δs PQT and RQU we have:

• QP = QR [Given]
• ∠PQT = ∠RQU [Given]
• ∠QPT = ∠QRU [From (ii)]

Therefore, by ASA congruence criterion, we have ΔQPT ≅ ΔQRU

⇒ QT = QU [cpct]

Note: Multiple other solved examples are available in the original document pages 23-34. Due to space constraints, only key examples are shown here. Students should refer to the complete document for all 24 illustrated examples.

Practice Questions:

Answer Key:

Long Answer Type Questions: