Construct a triangle ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct another triangle whose sides are`3/4` times the corresponding sides of ΔABC.

#### Solution

Following steps are involved in the construction of a ΔA'BC' whose sides are`3/4` of the corresponding sides of ΔABC:

**Step 1**

Draw a ΔABC with sides BC = 6 cm and AB = 5 cm and ∠ABC = 60°.

**Step 2**

Draw a ray BX making an acute angle with BC on the opposite side of vertex A.

**Step 3**

Locate 4 points (as 4 is greater in 3 and 4) B_{1}, B_{2}, B_{3 }and B_{4} on line segment BX.

**Step 4**

Join B_{4}C and draw a line through B_{3} parallel to B_{4}C intersecting BC at C'.

**Step 5**

Draw a line through C' parallel to AC intersecting AB at A'. ΔA'BC' is the required triangle.

**Justification**

The construction can be justified by proving

`A'B=3/4AB, BC'=3/4BC, A'C'=3/4AC`

In ΔA'BC' and ΔABC,

∠A'C'B = ∠ACB (Corresponding angles)

∠A'BC' = ∠ABC (Common)

∴ ΔA'BC' ∼ ΔABC (AA similarity criterion)

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

In ΔBB_{3}C' and ΔBB_{4}C,

∠B_{3}BC' = ∠B_{4}BC (Common)

∠BB_{3}C' = ∠BB_{4}C (Corresponding angles)

∴ ΔBB_{3}C' ∼ ΔBB_{4}C (AA similarity criterion)

`=>(BC')/(BC)=(BB_3)/(BB_4)`

`=>(BC')/(BC)=3/4 `

From (1) and (2), we obtain

`(A'B)/(AB)=(BC')/(BC)=(A'C')/(AC)=3/4`

`=>A'B=3/4AB, BC'=3/4BC, A'C'=3/4AC`

This justifies the construction.