Question

If ABC is an isosceles triangle in which AC = BC, AD and BE are respectively two altitudes to sides BC and AC, then prove that AE = BD.

Answer 1

In $\Delta ABC,AC=BC$ [given]
$\angle ABC=\angle CAB$
[angles opposite to equal sides are equal]
ie., $\angle ABD=\angle EAB...\left(i\right)$
In $\Delta AEBand\Delta BDA,$
$\angle AEB=\angle ADB={90}^{\circ }[given,AD\perp BCandBE\perp AC]\angle EAB=\angle ABD[fromEq.(i\left)\right]andAB=AB\left[commonside\right]\therefore \Delta AEB\cong \Delta BDA\left[byAAScongruencerule\right]\Rightarrow AE=BD\left[byCPCT\right]$