Draw a triangle ABC with BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are`4/5` times the corresponding sides of ΔABC.

#### Solution

Given:

∠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°

Following steps are involved in the construction of the required triangle:

**Step 1**

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

**Step 2**

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

**Step 3**

Locate 5 points (as 5 is greater in 5 and 4) B_{1, }B_{2, }B_{3, }B_{4 }and B_{5 }on BX such that BB_{1},_{ }B_{1}B_{2},_{ }B_{2}B_{3},_{ }B_{3}B_{4 }and B_{4}B_{5}.

**Step 4**

Join B_{5}C. Draw a line through B_{4} parallel to B_{5}C intersecting extended BC at C'.

**Step 5**

Through C', draw a line parallel to AC intersecting BA at A'.

ΔA'BC' is the required triangle.

**Justification**

The construction can be justified by proving that

`A'B=4/5AB,BC'=4/5BC andA'C,= 4/5AC`

In 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')`

In ΔBB_{5}C and ΔBB_{4}C',

∠B_{4}BC' = ∠B_{5}BC (Common)

∠BB_{4}C' = ∠BB_{5}C' (Corresponding angles)

∴ ΔBB_{4}C' ∼ ΔBB_{5}C (AA similarity criterion)

`(BC')/(BC)=(BB_4)/(BB_6)`

`=>(BC')/(BC)=4/5 `

Comparing (1) and (2), we obtain

`(A'B)/(AB)=(BC')/(BC)=(A'C')/(AC)=4/5`

This justifies the construction.