Sarthaks Test
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In Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

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Given:  AB = AC and CD || BA and AP and is the bisector of exterior 

∠CAD of ΔABC

To prove:

(i) ∠PAC = ∠BCA 

(ii) ABCP is a parallelogram.

Proof:

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