Question

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

Show that $\left(\mathrm{i}\right)\mathrm{\Delta APB}\cong \mathrm{\Delta CQD}$

$\left(\mathrm{ii}\right)\mathrm{AP}=\mathrm{CQ}$

Answer 1

Step $1:$ Proving $\mathrm{\Delta APB}\cong \mathrm{\Delta CQD}$:

In, $\mathrm{\Delta APB}$ and $\mathrm{\Delta CQD}$

$\begin{array}{rcl}\angle \mathrm{APB}& =& \angle \mathrm{CQD}(\mathrm{Each}\mathrm{equals}\mathrm{to}90°)\\ \angle \mathrm{ABP}& =& \angle \mathrm{CDQ}(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}\mathrm{for}\mathrm{AB}\parallel \mathrm{CD})\\ \mathrm{AB}& =& \mathrm{CD}(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}\mathrm{are}\mathrm{equal})\end{array}$

$\therefore △\mathrm{APB}\cong △\mathrm{CQD}(\mathrm{By}\mathrm{AAS}\mathrm{congruency}\mathrm{criteria})$

Step $2:$ Proving $\mathrm{AP}=\mathrm{CQ}$

$\therefore \mathrm{AP}=\mathrm{CQ}(\mathrm{By}\mathrm{CPCT})$

Hence proved.