Question

Given figure, $PA\perp AB,QB\perp AB$ and $PA=QB$, prove that $△OAP\cong OBQ$.

Answer 1

REF. Image.

Given :- PA $\perp$ AB $\Rightarrow \angle PAO={90}^{\circ }$

QB $\perp$ AB $\Rightarrow \angle OBQ={90}^{\circ }$

PA = QB

In $\Delta$ OAP and $\Delta$ OBQ.

$\angle$ APO = $\angle$ QBO [since both are ${90}^{\circ }]$

PA = QB [given]

$\angle$ POA = $\angle$ QBO [$\because$ vertically opposite angles are equal]

By ASA criterian concurrency

$\Delta$ OAP $\cong \Delta$ OBQ