$ \mathrm{ABCD} $ is a parallelogram and $ \mathrm{AP} $ and $ \mathrm{CQ} $ are perpendiculars from vertices $ \mathrm{A} $ and $ \mathrm{C} $ on diagonal $ \mathrm{BD} $ (see below figure). Show that
(i) $ \triangle \mathrm{APB} \cong \triangle \mathrm{CQD} $
(ii) $ \mathrm{AP}=\mathrm{CQ} $
"

Given :

$ABCD$ is a parallelogram and $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$.

To do :

We have to show that

(i) $\triangle {APB} \cong \triangle {CQD}$.
(ii) $AP = CQ$.

 Solution:

(i) In $\triangle ABP$ and $\triangle CDQ$,

$AB = CD$              [Opposite sides of parallelogram are equal]

$\angle APB = \angle CQD$      [Both are $90^o$ as given]

$\angle ABP = \angle CDQ$       [Alternate interior angles]

Therefore, $\triangle ABP \cong \triangle CDQ$.

(ii) $\triangle ABP \cong \triangle CDQ$.

So, all the angles and sides of both triangles are equal.

Therefore, $AP = CQ$.

Hence proved.

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Updated on: 10-Oct-2022

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