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.
Answers
Given : AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
To prove : (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.
Proof :
We have ,
AB = AC
∠C = ∠B ...............(1)
Now,
∠CAD = ∠B + ∠C
[Exterior angle theorem]
∠CAD = ∠C + ∠C
[From eq (1)]
2∠PAC = 2∠C
[AP is the bisector of exterior ∠CAD]
∠PAC = ∠C
[Alternate angles are equal ]
∠PAC= ∠BCA
Therefore, AP || BC
But, we have AB || CP
As we know that if both pairs of opposite sides are parallel of a quadrilateral then it is a parallelogram.
Hence ABCP is a parallelogram.
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