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In Fig, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that,(i) ∠PAC=∠BCA (ii) ABCP is a parallelogram |
Answer» Given, In figure, AB = AC CP || BA AP is bisector of exterior angle ∠CAD ∠C = ∠B Now, ∠CAD = ∠B + ∠C 2∠CAP = 2∠C ∠CAP = ∠C AP || BC But, AB || CP (given) Hence, ABCP is a parallelogram |
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