Question

In the given figure; $\angle BAC={90}^{\circ }$ and $AD\perp BC.$ Show that $\Delta ADC$ and $\Delta BAC$ are similar.

Answer 1

In $△ADC$ and $△BAC$,

$\angle ADC=\angle BAC$ (Each ${90}^{\circ }$)

$\angle DCA=\angle BCA$ (Common)

$\angle DAC=\angle ABC$ (Third angle)

Thus, $△ADC\sim △BAC$ (AAA rule)