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In the figure, $ABCD$ is a parallelogram in which $\angle A = 60^o$. If the bisectors of $\angle A$ and $\angle B$ meet at $P$, prove that $AD = DP, PC= BC$ and $DC = 2AD$.
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Academic Mathematics NCERT Class 9

Given:

$ABCD$ is a parallelogram in which $\angle A = 60^o$.

The bisectors of $\angle A$ and $\angle B$ meet at $P$.
To do:

We have to prove that $AD = DP, PC= BC$ and $DC = 2AD$.

Solution:

$\angle A + \angle B = 180^o$

$60^o + \angle B = 180^o$

$\angle B = 180^o - 60^o = 120^o$

$DC \parallel AB$

This implies,

$\angle PAB = \angle DPA$ (Alternate angles)

$\angle PAD = \angle DPA$ (Since $\angle PAB = \angle PAD$)

Therefore,

$AB = DP$ (Sides opposite to equal angles are equal)

Similarly,

$\angle PBC = \angle PCB$ ($\angle PAB = \angle BCA$)

Therefore,

$PC = BC$

$DC = DP + PC$

$= AD + BC$

$= AD + AB$

$= 2AB$ (Sinve $BC = AD$)

Hence, $DC = 2AD $.

Updated on: 10-Oct-2022

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