Question

In the given figure, $△ABC$ is an isosceles right angled triangle with $\angle B=90$ such that $PQ\perp AC,ST\perp AC$ where P lies on AB and S lies on BC. Then prove that $△AQP\sim △CTS$

Answer 1

$△ABC$ is an isosceles right angled triangle

$⟹\angle A={45}^{\circ },\angle B={90}^{\circ },\angle C={45}^{\circ }$

In $△AQP,\angle AQP={90}^{\circ },\angle QAP={45}^{\circ },\angle APQ={180}^{\circ }-{90}^{\circ }-{45}^{\circ }={45}^{\circ }$

In $△CTS,\angle CTS={90}^{\circ },\angle TCS={45}^{\circ },\angle CST={180}^{\circ }-{90}^{\circ }-{45}^{\circ }={45}^{\circ }$

In $△AQP,△CTS$

$\angle AQP=\angle CTS,\angle QAP=\angle TCS,\angle APQ=\angle CST$

By $AAA$ congruency ,$△AQP\sim △CTS$