Question

ABC is a triangle. The bisector of the exterior angle at B and the bisector of angle C intersect each other at D.Prove angle BDC =1/2 angle A.​

Answer 1

ABC is a triangle. The bisectors of the internal angle ∠B and external angle ∠C intersect at D. If ∠BDC = 60, then ∠A is?

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The problem can be solved based on the theorem:

“External angle is equal to the sum of two internal opposite angles.”

Applying the above theorem to triangle BDC,

∠C2=∠B2+∠BDC

Given ∠BDC=60°,

∠C2=∠B2+60°⋯(1)

Applying the theorem to triangle ABC,

∠C=∠A+∠B⟹∠C2=∠A2+∠B2⋯(2)

From equations (1) and (2), we get

∠B2+60°=∠A2+∠B2=∠C2

⟹∠B2+60°=∠A2+∠B2

⟹60°=∠A2⟹∠A=120°.

by shatakshi anand

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