In ∆PAB,∠PBD = ∠PAB + ∠APB Exterior angle theorem⇒∠APB = ∠PBD - ∠PAB 1In ∆ABC,∠CBD = ∠CAB + ∠ACB Exterior angle theorem⇒∠ACB = ∠CBD - ∠CAB 2⇒∠ACB = 2∠PBD - 2∠PAB As, AP bisects ∠A and ∠BP bisects ∠CBD⇒∠ACB = 2∠APB Using 1⇒∠APB = 12×80° = 40°

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In ∆ PAB, ∠ PBD = ∠ PAB + ∠ APB (Exterior angle theorem) \Rightarrow ∠ APB = ∠ PBD - ∠ PAB . . . . (1) In ∆ ABC, ∠ CBD = ∠ CAB + ∠ ACB (Exterior angle theorem) \Rightarrow ∠ ACB = ∠ CBD - ∠ CAB . . . . . (2) \Rightarrow ∠ ACB = 2 ∠ PBD - 2 ∠ PAB (As, AP bisects ∠ A and ∠ BP bisects ∠ CBD) \Rightarrow ∠ ACB = 2 ∠ APB [Using (1)] \Rightarrow ∠ APB = \frac{1}{2} \times 80 ° = 40 °

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