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Question
In a triangle ABC. If D is a point on BC such that ∠CAD = ∠B, then prove that: ∠ADC = ∠BAC.
Solution
Given,∠CAD = ∠B ...(i)
By exterior angle property,
∠ADB = ∠CAD + ∠C
Also, ∠ADC = ∠BAD + ∠B
⇒ ∠ADC = ∠BAD + ∠CAD ....[From (i)]
⇒ ∠ADC = ∠BAC.
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