Construct a triangle similar to a given triangle PQR with its sides equal to `2/3` of the corresponding sides of the triangle PQR (scale factor `2/3 < 1`)
Given ∆PQR, we are required to construct another triangle whose sides are `2/3` of the corresponding sides of the ∆PQR
Steps of construction:
(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q1, Q2 and Q3 on QX.
So that QQ1 = Q1Q2 = Q2Q3
(iv) Join Q3 R and draw a line through Q2 parallel to Q3 R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆P’QR’ is the required triangle.