# 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"

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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.

Updated on 10-Oct-2022 13:41:01