Question

Given $\overline{AB}\cong \overline{AC}$, $m\angle CDB=m\angle BEC=90$. Prove that $\overline{AD}=\overline{AE}$

Answer 1

In $\Delta AEB$ & $\Delta ADC$

$AB=AC$ (given)

$\angle BAE=\angle CAD$ (common)

$\angle AEB=\angle ADC$ (both ${90}^o$)

$\therefore \Delta AEB\cong \Delta ADC\left(AAS\right)$

$\therefore$ by $CPCT\to AD=AE$