Question

In the adjacent figure, CD and BE are altitudes of an isosceles triangle ABC with $AC=AB$. Prove that $AE=AD$

Answer 1

Given $DC$ and $BE$ are altitude and $AB=AC$

So Here $\angle CDB=\angle CDA={90}^{\circ }$ --------------- $\left(1\right)$

And$\angle BEC=\angle BEA={90}^{\circ }$ --------------- $\left(2\right)$

So from equation $\left(1\right)$ and $\left(2\right)$

we can say

$\angle CDB=\angle BEC{90}^{\circ }$ --------------- $\left(3\right)$

And As given $ABC$ is a isosceles triangle so , from base angle theorem,

we can say that

$\angle ABC=\angle ACB$ ---------- $\left(4\right)$

Now In $△CBD$ and $△BCE$

$\angle CDB=\angle BEC$ ( From equation $\left(3\right)$

$\angle DBC=\angle ECB$ ( As $\angle ABC=\angle DBC$) ( same angles )

And $\angle ACB=\angle ECB$( same angles )

And from equation $\left(4\right)$ we know $\angle ABC=\angle ACB$

And $BC=BC$ ( Common side )

Hence,$△CBD$ is congruent to $△BCE$ ( By AAS rule )

So,$AE=AD$ ( By CPCT rule ) ( Hence proved )