Construct an isosceles
triangle whose base is 8 cm and altitude 4 cm and then another
triangle whose side are **
**times the corresponding sides of the isosceles triangle.

Give the justification of the construction.

Let us assume that ΔABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.

A ΔAB'C' whose sides are times of ΔABC can be drawn as follows.

**Step 1**

Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.

**Step 2**

Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.

**Step 3**

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

**Step 4**

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

**Step 5**

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

**Step 6**

Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle.

**Justification**

The construction can be justified by proving that

In ΔABC and ΔAB'C',

∠ABC = ∠AB'C' (Corresponding angles)

∠BAC = ∠B'AC' (Common)

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

… (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)

On comparing equations (1) and (2), we obtain

⇒

This justifies the construction.

**
**