Question:
In figure, 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.
Solution:
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]
Therefore, ABCP is a parallelogram.