#### Question

Construct a triangle similar to a given Δ*ABC* such that each of its sides is (2/3)^{rd} of the corresponding sides of Δ*ABC*. It is given that BC = 6 cm, ∠*B* = 50° and ∠*C* = 60°.

#### Solution

Given that

Construct a triangle of given data, BC = 6 cm, ∠*B* = 50° and ∠*C* = 60° and then a triangle similar to it whose sides are (2/3)^{rd} of the corresponding sides of Δ*ABC*.

We follow the following steps to construct the given

Step of construction

Step: I- First of all we draw a line segment Bc = 60°.

Step: II- With *B *as centre draw an angle ∠*B* = 50°.

Step: III- With *C *as centre draw an angle ∠*C* = 60° which intersecting the line drawn in step II at *A.*

Step: IV- Joins *AB *and A*C *to obtain Δ*ABC*.

Step: V -Below *BC, *makes an acute angle ∠CBX = 60°.

Step: VI -Along *BX,* mark off three points B_{1}, B_{2} and B_{3} such that BB_{1} = B_{1}B_{2} = B_{2}B_{3}

Step: VII -Join B_{3}C.

Step: VIII -Since we have to construct a triangle each of whose sides is two-third of the corresponding sides of Δ*ABC*.

So, we take two parts out of three equal parts on *BX *from point B_{2} draw B_{2}C' || B_{3}C, and meeting *BC *at *C’.*

Step: IX -From *C’ *draw C'A' || AC and meeting *AB *at A*’*

Thus, ΔA'BC' is the required triangle, each of whose sides is two third of the corresponding sides of Δ*ABC**.*