$ABC$ is an isosceles triangle in which $AC = BC$. $AD$ and $BE$ are respectively two altitudes to sides $BC$ and $AC$. Prove that $AE = BD$.

(N/A) In $\triangle ADC$ and $\triangle BEC$ we have:
$AC = BC$ [Given] ... $(1)$
$\angle ADC = \angle BEC = 90^{\circ}$ [Given as altitudes]
$\angle ACD = \angle BCE$ [Common angle]
Therefore,$\triangle ADC \cong \triangle BEC$ [By $AAS$ congruence rule]
Therefore,$CD = CE$ ... $(2)$ [$CPCT$]
Subtracting $(2)$ from $(1)$,we get:
$AC - CE = BC - CD$
$AE = BD$
Hence,proved.