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
2∠CAP = 2∠C
∠CAP = ∠C
AP ⎸⎸BC
But, AB ⎸⎸CP (given)
Hence ABCP is a parallelogram