In Fig. 14.98, 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.

Given,


In figure 14.98


AB = AC


CP ⎸⎸BA


AP is bisector of exterior angle CAD


AB = AC


C = B


NOW,


CAD = B +C


2CAP = 2C


CAP = C


AP ⎸⎸BC


But, AB ⎸⎸CP (given)


Hence ABCP is a parallelogram


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