Construction

Geometric construction is a fundamental skill in mathematics that involves creating precise geometric figures using only basic tools such as a compass, straightedge, and pencil. This ancient art, dating back to the Greek mathematicians, forms the foundation of many mathematical concepts and practical applications in engineering, architecture, and design.

Key Concepts Covered:

• Division of line segments in a given ratio using two different methods
• Construction of similar triangles with specified scale factors
• Construction of tangents to circles from external and internal points
• Application of the Basic Proportionality Theorem

Division of a Line Segment

Definition

The division of a line segment internally in a given ratio m:n means finding a point P on the line segment AB such that the ratio of AP to PB equals m:n, where both m and n are positive integers.

When we divide a line segment AB internally in the ratio m:n, we are essentially finding a point P on AB such that AP:PB = m:n. This construction has numerous practical applications in architecture, engineering design, and scale modeling.

Method 1: Standard Method for Line Segment Division

Steps of Construction:

1. Step 1: Draw a line segment AB of given length using a ruler. Mark the endpoints clearly as points A and B.
2. Step 2: Draw a ray AX making an acute angle with AB. The angle should be less than 90 degrees to ensure proper construction and visibility of subsequent points.
3. Step 3: Along ray AX, mark off (m + n) points A₁, A₂, A₃, ..., A_m+n such that AA₁ = A₁A₂ = A₂A₃ = ... = A_m+n-1A_m+n. These points must be equally spaced along the ray.
4. Step 4: Join point B to the last point A_m+n on ray AX. This creates a line segment BA_m+n.
5. Step 5: Through the point A_m (the m-th point), draw a line parallel to A_m+nB by making an angle equal to ∠AA_m+nB at A_m. Suppose this parallel line meets AB at point P.

Figure 1: Division of a line segment in a given ratio using the standard method

Result: The point P so obtained is the required point which divides AB internally in the ratio m:n.

Example: Dividing a Line Segment in the Ratio 3:2

Problem Statement:

Given a line segment AB, divide it in the ratio 3:2.

Detailed Solution:

1. Draw any ray AX, making an acute angle with AB. This angle should be convenient for construction, typically between 30° and 60°.
2. Locate 5 (= m + n = 3 + 2) points A₁, A₂, A₃, A₄, and A₅ on AX such that:

   AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅

   Use a compass to ensure equal spacing between consecutive points.
3. Join point B to A₅, creating line segment BA₅.
4. Through point A₃ (since m = 3), draw a line parallel to A₅B by making an angle equal to ∠AA₅B. This parallel line intersects AB at point C.

Conclusion: Point C divides AB such that AC:CB = 3:2

Mathematical Justification

The construction works based on the Basic Proportionality Theorem (also known as Thales' Theorem). Since the line through A₃ is parallel to A₅B, we can apply this theorem to triangles formed by these lines.

By Basic Proportionality Theorem: AA₃/A₃A₅ = AC/CB

Since by construction, AA₃ contains 3 equal segments and A₃A₅ contains 2 equal segments, we have:

AC/CB = 3/2

This proves that point C divides AB in the ratio 3:2.

Method 2: Alternative Method for Line Segment Division

This alternative method provides a different approach to dividing a line segment internally in a given ratio m:n. This method is particularly useful when working with certain geometric constraints or when a symmetrical construction is preferred.

Steps of Construction:

1. Step 1: Draw a line segment AB of given length using a ruler.
2. Step 2: Draw any ray AX making an acute angle ∠BAX with AB.
3. Step 3: Draw a ray BY on the opposite side of AX, parallel to AX, making an angle ∠ABY equal to ∠BAX. This creates a symmetrical configuration.
4. Step 4: Mark off m points A₁, A₂, ..., A_m on ray AX and n points B₁, B₂, ..., B_n on ray BY such that:

   AA₁ = A₁A₂ = ... = A_m-1A_m = BB₁ = B₁B₂ = ... = B_n-1B_n

5. Step 5: Join A_mB_n. Suppose it intersects AB at point P.

Result: Point P is the required point dividing AB in the ratio m:n.

Alternative method for division of a line segment internally in a given ratio

Use the following steps to divide a given line segment AB internally in a given ration m : n, where m and natural members.

STEPS OF CONSTRUCTION:

(i) Draw a line segment AB of given length.

(ii) Draw any ray AX making an acute angle ∠BAX with AB.

(iii) Draw a ray BY, on opposite side of AX, parallel to AX making an angle ∠ABY equal to ∠BAX.

(iv) Mark off a points A₁, A₂,...,Aₘ, on AX and n points B₁, B₂,...,Bₙ on BY such that AA₁ = A₁A₂ = ...... = Aₘ₋₁Aₘ = B₁B₂ = ....Bₙ₋₁Bₙ.

(v) Join AₘBₙ. Suppose it intersect AB at P.

• A horizontal line segment AB with points A and B marked

• Ray AX extending upward from A with points A₁, A₂, A₃, Aₘ₋₁, Aₘ, and X marked along it

• Ray BY extending downward from B with points B₁, B₂, Bₙ, and Y marked along it

• Point P where line AₘBₙ intersects AB

• Labels showing "m" and "n" segments on line AB]

The point P is the required point dividing AB in the ratio m : n.

This geometric construction method allows division of a line segment in any given ratio m:n by creating equal divisions on two parallel rays and connecting specific points to find the division point on the original segment.

Example. Decide a line segment of length 6 cm internally in the ratio 3:4.

Sol. Follow the following steps:

STEPS OF CONSTRUCTION:
(i) Draw a line segment AB of length 6 cm.
(ii) Draw any ray AX making an acute angle ∠BAX with AB.
(iii) Draw a ray BY parallel to AX by making ∠ABY equal to ∠BAX.
(iv) Mark of three points A₁, A₂, A₃ on AX and 4 points B₁, B₂, B₃, B₄ on BY such that AA₁ = A₁A₂ = A₂A₃ = BB₁ = B₁B₂ = B₂B₃ = B₃B₄.
(v) Join A₃B₄. Suppose it intersects AB at a point P. Then, P is the point dividing AB internally in the ratio 3:4.

Construction of A triangle similar to a given triangle

Scale Factor:
The ratio of the sides of the triangle to be constructed with the corresponding sides of the given triangle is known as their scale factor.

Steps of Construction when m < n:

1. Construct the given triangle ABC by using the given data.

2. Take any one of the three sides of the given triangle as the base. Let AB be the base of the given triangle.

3. At one end, say A, of the base AB, construct an acute angle ∠BAX∠BAX below the base AB.

4. Along AX, mark n points A1,A2,A3,…,AnA1,A2,A3,…,An such that AA1=A1A2=…=An−1AnAA1=A1A2=…=An−1An.

5. Join AnAn to B.

6. Draw A_mB′ parallel to A_nB which meets AB at B′.

7. From B′ draw B′C′∥CB meeting AC at C′.

Triangle AB′C′ is the required triangle each of whose side is (m/n)^th of the corresponding side of △ABC.

Ex.: Construction a △ABC in which AB = 5cm, BC = 6cm and AC = 7cm. Now, construct a triangle similar to △ABC such that each of its side is two-third of the corresponding side of △ABC.

Sol. Steps of Construction
(i) Draw a line segment AB = 5 cm.
(ii) With AA as centre and radius AC=7cm, draw an arc.
(iii) With BB as centre and BC = 6cm, draw another arc, intersecting the arc drawn in step (ii) at C.
(iv) Join AC and BC to obtain △ABC.
(v) Below ABAB, make an acute angle ∠BAX.
(vi) Along AX, mark off three points (greater of 2 and 3 in 2/3) A₁,A₂,A₃ such that A₁A₂ = A₂A₃.

(vii) Join A₃B.

(viii) Draw A₂B' ∥ A₃B, meeting AB at B'.

(iv) From B', draw B'C' ∥ BC, meeting AC at C'.

AB'C' is the required triangle, each of the whose sides is two-third of the corresponding sides of △ABC.

Steps of Construction when m > n:

(i) Construct the given triangle by using the given data.

(ii) Take any of the three sides of the given triangle and consider it as the base. Let AB be the base of the given triangle.

Here is the text conversion of the image with all mathematical symbols and expressions:

(iii) At one end, say A, of base AB construct an acute angle ∠BAX below base AB i.e. on the composite side of the vertex C.

(iv) Along AX, mark-off m (large of m and n) points A₁, A₂,.....,Aₘ on AX such that AA₁ = A₁A₂ = .... Aₘ₋₁Aₘ.

(v) Join Aₙ to B and draw a line through Aₘ parallel to Aₙ B, intersecting the extended line segment AB at B'.

(vi) Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.

(vii) ΔABC' so obtained is the required triangle.

[The image shows a geometric construction with:

• Triangle ABC (original triangle with vertices A, B, C)
• Point X on a ray from A below the base AB
• Points A₁, A₂, Aₙ, Aₘ marked sequentially along AX
• Point B' on the extension of AB
• Point C' on the extension of AC
• Dotted lines showing the parallel constructions from Aₘ to B' and from B' to C'
• The final triangle AB'C' is the enlarged triangle]

Construction of Tangents from a point outside the circle

When centre is given:

⇒ When point of tangency is on the circle.

Given: A circle with centre O.

Required: To draw a tangent from point P on the circle.

Steps of construction:

1. Take a point O on the plane of the paper and draw a circle of given radius.
2. Take a point P on the circle.
3. Join OP.
4. Construct ∠OPX = 90°.
5. Produce XP to Y to get XPY as the required tangent.

[DIAGRAM showing a circle with center O, point P on the circle, radius OP drawn vertically, and a horizontal tangent line XPY passing through P perpendicular to OP]

⇒ When point of tangency is outside the circle

Given: A circle with centre O.

Required: To draw a tangent from an external point, i.e. P.

Steps of construction:

1. Join the centre O of the circle to the given external point, i.e. P.
2. Draw ⊥ bisector of OP, intersecting OP at O′
3. Taking O′ as centre and OO′ = PO′ as radius, draw a circle to intersect the given circle at T and T′
4. Join PT and PT′ to get the required tangents as PT and PT′.

WHEN CENTRE IS NOT GIVEN:

⇒ When point of tangency is on the circles.

Given: A circle and a point P on it.

Required: To draw a tangent at P without using centre of the circle.

Steps of construction:

1. Draw any chord PQ of the circle through P as in figure.
2. Take any point R on the major arc PQ and join PR and QR.
3. Construct ∠QPX equal to ∠PRQ.

[THIS IS FIGURE: A circle with points P, Q, and R marked. R is on the major arc. Lines PR and QR are drawn forming a triangle. A line PX extends from P as a tangent to the circle.]

Then PX is the required tangent at P to the circle.

⇒ When point of tangency is outside the circle.

Given: A circle and a point P outside it.

Required: To draw a tangent from point P without using the centre.

Steps of construction:

1. Let P be the external point from where the tangents are to be drawn to the given circle.
2. Through P draw a secant PAB to intersect the circle at A and B.
3. Produce AP to a point C such that AP = PC.
4. Draw a semi-circle with BC as diameter.
5. Draw PD ⊥ CB, intersecting the semi-circle at D.
6. With P as centre and PD as radius draw arcs to intersect the given circle at T and T'.
7. Joint PT and PT'. PT and PT' are the required tangents.

Ex.: Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

Given:

(i) A circle with centre O and radius 6 cm.

(ii) A point P at a distance of 10 cm from O.

Required:

To draw two tangents to the circle from point P and measure the lengths of the tangents.

Steps of construction:

1. Bisect the line-segment OP. Let the point of bisection be M.
2. Taking M as centre and OM as radius, draw a circle. Let is intersect the circle at the points Q and R.
3. Join PQ and PR. These are the required tangents. On measuring, PQ = PR = 8 cm.

Diagram:

The diagram shows:

• A circle with centre O and radius 6 cm
• Point P at distance 10 cm from O
• Point M is the midpoint of OP
• A larger circle with centre M and radius OM
• Points Q and R where the larger circle intersects the original circle
• Tangent lines PQ and PR drawn from P to the circle
• The measurements show OP = 10 cm and the radius = 6 cm
• The tangent lengths PQ = PR = 8 cm

Ex.: Draw tangents from an external point P to a circle of radius 4 cm without using the centre.

Sol. Given: A circle of radius 4 cm.

Required: To draw two tangents from an external point P.

Steps of construction:

1. Draw a circle of radius 4 cm.
2. Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B.
3. Produce AP to C such that PA = CP.
4. Draw a semi-circle with CB as diameter.
5. Draw PD ⊥ CB, intersecting the semi-circle at D.
6. With P as centre and PD as radius draw arcs to intersect the given circle at T and T′.
7. Join PT and PT′. Then PT and PT′ are the required tangents.

In this comprehensive guide, we will explore the systematic methods for dividing line segments in given ratios and constructing triangles similar to given triangles. These construction techniques are essential for understanding proportionality, similarity, and the fundamental principles of Euclidean geometry.