Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides are `2/3` of the corresponding sides of the first triangle. Give the justification of the construction.

Construct a triangle with sides 5 cm, 4 cm and 6 cm. Then construct another triangle whose sides are `2/3` times the corresponding sides of first triangle.

#### Solution

**Step 1**

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

**Step 2**

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

**Step 3**

Locate 3 points A_{1}, A_{2}, A_{3} (as 3 is greater between 2 and 3) on line AX such that AA_{1 }= A_{1}A_{2} = A_{2}A_{3}.

**Step 4**

Join BA_{3} and draw a line through A_{2 }parallel to BA_{3} to intersect AB at point B'.

**Step 5**

Draw a line through B' parallel to the line BC to intersect AC at C'.

ΔAB'C' is the required triangle.

**Justification**

The construction can be justified by proving that

`AB' = 2/3AB, B'C' = 2/3BC, AC' = 2/3 AC`

By construction, we have B’C’ || BC

∴ ∠AB'C'= ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,

∠AB'C' = ∠ABC (Proved above)

∠B'AC' = ∠BAC (Proved above)

∴ ΔAB'C' ~ ΔABC (AA similarity criterion)

`=> (AB')/(AB) = (B'C')/(BC) = (AC')/(AC) ....(1)`

In ΔAA_{2}B' and ΔAA_{3}B,

∠A_{2}AB' = ∠A_{3}AB (Common)

∠AA_{2}B' = ∠AA_{3}B (Corresponding angles)

∴ ΔAA_{2}B' ∼ ΔAA_{3}B (AA similarity criterion)

`=> (AB')/(AB) = (`

`=> (AB')/(AB) = 2/3 ....(2)`

From equations (1) and (2), we obtain

`(AB')/(AB) = (B'C')/(BC) = (AC')/(AC) = 2/3`

`=>AB' = 2/3(AB), B'C' = 2/3(BC), AC' = 2/3(AC)`

This justifies the construction.