Question:
In a triangle ABC, AD bisects ∠A and ∠C > ∠B. Prove that ∠ADB > ∠ADC.
Solution:
∵ ∠C > ∠B [Given]
⇒ ∠C + x > ∠B + x [Adding x on both sides]
⇒ 180° - ∠ ADC > 180° - ∠ADB
⇒ - ∠ ADC > - ∠ADB
⇒ ∠ADB > ∠ADC
Hence proved.
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