Question

14. In the adjoining figure, AB = AC, CPBA and AP is the bisector of ZCAD.
Prove that:
(i) ZPAC = ZBCA
(ii) ABCP is a parallelogram.
[Hint: ZCAD = ZABC + ZACB = 2 ZACB (Why)?And so,ZPAC=1/2ZCAD =ZACB.
ZPAC = AP|| BC]
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Answer 1

Answer:

where is the figure bro????

Answer 2

you have forgotten to send the picture