Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are `4/3 `times the corresponding side of ΔABC. Give the justification of the construction.

#### Solution

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

**Step 1**

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

**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 4 and 3), B_{1, }B_{2, }B_{3, }B_{4}, on BX.

**Step 4**

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

**Step 5**

Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.

**Justification**

The construction can be justified by proving that

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

n ΔABC and ΔA'BC',

∠ABC = ∠A'BC' (Common)

∠ACB = ∠A'C'B (Corresponding angles)

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

`=>(AB)/(A'B) = (BC)/(BC') = (AC)/(A'C') ....1`

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') = `

`=>(BC)/(BC') = 3/4 ...(2)`

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

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

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

This justifies the construction.