Question

In Fig. below, 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

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

∠CAD of ΔABC

To prove:

i)   ∠PAC = ∠BCA

ii)  ABCP is a parallelogram

Proof:

i) We have,

AB = AC

⟹ ∠ACB = ∠ABC              [Opposite angles of equal sides of triangle are equal]

Now, ∠CAD = ∠ABC + ∠ACB

 ⇒ ∠PAC + ∠PAD  = 2∠ACB    (∵ ∠PAC = ∠PAD )

⇒  2∠PAC  = 2∠ACB

⇒ ∠PAC = ∠ACB

ii) Now,

∠PAC = ∠BCA

 ⇒ AP || BC

And, CP || BA                [Given]

∴ ABCP is a parallelogram .