Question

In a triangle ABC. If D is a point on BC such that ∠CAD = ∠B, then prove that: ∠ADC = ∠BAC.

Answer 1

Given,∠CAD = ∠B                 ...(i)
By exterior angle property,
∠ADB = ∠CAD + ∠C
Also, ∠ADC = ∠BAD + ∠B
⇒ ∠ADC = ∠BAD + ∠CAD      ....[From (i)]
⇒ ∠ADC = ∠BAC.