Question

# $ABCD$ is a parallelogram and $AP$ and$CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$. Show that $\left(\mathrm{i}\right)\mathrm{\Delta APB}\cong \mathrm{\Delta CQD}$ $\left(\mathrm{ii}\right)\mathrm{AP}=\mathrm{CQ}$

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Solution

## Step $1:$ Drawing the diagram:$ABCD$ is a parallelogram$AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$.Step $2:$ Proving $\mathrm{\Delta APB}\cong \mathrm{\Delta CQD}$:In, $\mathrm{\Delta APB}$ and $\mathrm{\Delta CQD}$$\begin{array}{rcl}\angle \mathrm{APB}& =& \angle \mathrm{CQD}\left(\mathrm{Each}\mathrm{equals}\mathrm{to}90°\right)\\ \angle \mathrm{ABP}& =& \angle \mathrm{CDQ}\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}\mathrm{for}\mathrm{AB}\parallel \mathrm{CD}\right)\\ \mathrm{AB}& =& \mathrm{CD}\left(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}\mathrm{are}\mathrm{equal}\right)\end{array}$$\therefore △\mathrm{APB}\cong △\mathrm{CQD}\left(\mathrm{By}\mathrm{AAS}\mathrm{congruency}\mathrm{criteria}\right)$Step $3:$ Proving $\mathrm{AP}=\mathrm{CQ}$$\therefore \mathrm{AP}=\mathrm{CQ}\left(\mathrm{By}\mathrm{CPCT}\right)$Hence proved.

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