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