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# In parallelogram $ \mathrm{ABCD} $, two points $ \mathrm{P} $ and $ \mathrm{Q} $ are taken on diagonal $ \mathrm{BD} $ such that $ \mathrm{DP}=\mathrm{BQ} $ (see below figure). Show that:

**(i)** $ \triangle \mathrm{APD} \equiv \triangle \mathrm{CQB} $

**(ii)** $ \mathrm{AP}=\mathrm{CQ} $

**(iii)** $ \triangle \mathrm{AQB} \equiv \triangle \mathrm{CPD} $

**(iv)** $ \mathrm{AQ}=\mathrm{CP} $

**(v)** $ \mathrm{APCQ} $ is a parallelogram

"

Given:

In parallelogram \( \mathrm{ABCD} \), two points \( \mathrm{P} \) and \( \mathrm{Q} \) are taken on diagonal \( \mathrm{BD} \) such that \( \mathrm{DP}=\mathrm{BQ} \)

To do :

We have to show that

(i) \( \triangle \mathrm{APD} \equiv \triangle \mathrm{CQB} \)

(ii) \( \mathrm{AP}=\mathrm{CQ} \)

(iii) \( \triangle \mathrm{AQB} \equiv \triangle \mathrm{CPD} \)

(iv) \( \mathrm{AQ}=\mathrm{CP} \)

(v) \( \mathrm{APCQ} \) is a parallelogram

Solution :

(i) In $\triangle APD$ and $\triangle CQB$,

$DP = BQ$ (Given)

$\angle ADP = \angle CBQ$ (Alternate interior angles are equal)

$AD = BC$ (Opposite sides of a parallelogram are equal)

Therefore, by SAS congruency,

$\triangle APD \cong \triangle CQB$

(ii) $\triangle APD \cong \triangle CQB$

This implies,

$AP = CQ$ (CPCT)

(iii) In $\triangle AQB$ and $\triangle DPC$,

$BQ = DP$ (Given)

$\angle ABQ = \angle CDP$ (Alternate interior angles are equal)

$AB = CD$ (Opposite sides of a parallelogram are equal)

Therefore, by SAS congruency,

$\triangle AQB \cong \triangle CPD$

(iv) $\triangle AQB \cong \triangle CPD$

This implies,

$AQ = CP$ (CPCT)

(v) $AP = CQ$

$AQ = CP$

This implies,

In quadrilateral $APCQ$ opposite sides are equal and opposite angles are equal.

Therefore, $APCQ$ is a parallelogram.

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