In the figure, ABC is an isosceles triangle in
which AB = AC. The side BA is produced to D
and CP || BA is drawn. The bisector of angle CAD
cuts CP at P. Prove that ABCP is a parallelogram.
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angle BAC = angle ACP alternate angle (Since BA || CP And CA is transversal)
Since AB=AC angle ABC = angle ACB
angle CAD = angle ABC + angle ACB
2*angle ABC = angle CAD
angle ABC = 1/2 angle CAD= angle CAP
Now we can prove ∆ABC and ∆ ACP are congruence (A-S-A)
So AB=CP
AP= BC
AP || BC
Hence ABCP is a Parallelogram
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ABCP is a parallelogram(proved)
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