Question

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 1

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