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.
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