Question

In the given figure ,ABC is a triangle with ∠B = 2 ∠C.

D is a point on BC such that AD bisects ∠BAC and AD=CD.

Prove that ∠BAC = 72°.​

Answer 1

Hi,

Explanation:

`Let's\ assume\ \angle{C}=\alpha\\\\\angle{B}=2*\angle{C}=2\alpha\\\\\angle{DAC}=\angle{C}=\alpha\\AD\ bisects\ \angle{BAC}\\\angle{BAC}=2*\alpha\\\\\angle{B}+\angle{A}+\angle{C}=2*\alpha+2*\alpha+\alpha=5*\alpha=180^o\\\\\alpha=36^o\\2*\alpha=2*36^o=72^o\\`