Draw a triangle ABC with side BC=6cm,AB=5cm & angle ABC=60' . Then construct a triangle with sides 3/4 of the corresponding sides of the triangle ABC.
A ΔA'BC' whose sides are of the corresponding sides of ΔABC can be drawn as follows.
Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
Locate 4 points (as 4 is greater in 3 and 4), B1, B2, B3, B4, on line segment BX.
Join B4C and draw a line through B3, parallel to B4C intersecting BC at C'.
Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.
The construction can be justified by proving
In ΔA'BC' and ΔABC,
∠A'C'B = ∠ACB (Corresponding angles)
∠A'BC' = ∠ABC (Common)
∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)
In ΔBB3C' and ΔBB4C,
∠B3BC' = ∠B4BC (Common)
∠BB3C' = ∠BB4C (Corresponding angles)
∴ ΔBB3C' ∼ ΔBB4C (AA similarity criterion)
From equations (1) and (2), we obtain
This justifies the construction.