Question -
Answer -
In ∆ABC, ∠B = 35°, ∠C = 65° and AP is the bisector of ∠BAC which meets BC in P.
Arrange PA, PB and PC in descending order In ∆ABC,
∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
⇒ ∠A + 35° + 65° = 180°
∠A + 100°= 180°
∴ ∠A =180°- 100° = 80°
∵ PA is a bisector of ∠BAC
∴ ∠1 = ∠2 =
= 40° Now in ∆ACP, ∠ACP > ∠CAP
⇒ ∠C > ∠2
∴ AP > CP …(i)
Similarly, in ∆ABP,
∠BAP > ∠ABP ⇒ ∠1 > ∠B
∴ BP > AP …(ii)
From (i) and (ii)
BP > AP > CP