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
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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
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
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