Question

In Figure, BP bisects ∠ABC and AB = AC. Find x.

Answer 1

In the figure,

AB = AC, and BP bisects ∠ABC

AP || BC is drawn.

Now ∠PBC = ∠PBA .........(∵ PB is the bisector of ∠ABC)

∵ AP || BC

∴ ∠APB = ∠PBC ............(Alternate angles)

⇒ x = ∠PBC ..................(i)

In Δ ABC, ∠A = 60°
and ∠B = ∠C ...................(∵ AB = AC)
But ∠A + ∠B + ∠C = 180°.....................(Angles of a triangle)

⇒ 60° + ∠B + ∠C = 180°

⇒ 60° + ∠B + ∠B = 180°

⇒ 2 ∠B = 180° − 60° = 120°

∴ ∠B =`(120°)/2=60°`

⇒ `1/2∠"B"=(60°)/2=30°`

⇒ ∠PBC = 30°

∴ From (i),

x = 30°