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

Dear Student,

Here is the solution of your asked query:

Given  AD and BE are altitude and AC =  BC  , So we have our figure , As  :

Here

$\angle$ BEA  =  $\angle$ BEC  = 90$°$  --------------- ( 1 )

And
$\angle$ ADB  =  $\angle$ ADC = 90$°$  ---------------- ( 2 )
So from equation 1 and 2 , we can say

$\angle$ BEA  =  $\angle$ ADB  = 90$°$  ---------------- ( 3 )

And As given ABC is a isosceles triangles so , from base angle theorem ,we can say that

$\angle$ CAB  =  $\angle$ CBA  ----------------- ( 4 )

Now In $∆$ BAE  and $∆$ ABD 

$\angle$ BEA  =  $\angle$ ADB  ( From equation 3 )

$\angle$ EAB  =  $\angle$ DBA  ( As $\angle$ CAB  = $\angle$ EAB ( same angles )  And  $\angle$ CBA =  $\angle$ DBA ( same angles )  And from equation 4 we know $\angle$ CAB  =  $\angle$ CBA )
And
AB  =  AB  ( Common side  )

Hence

$∆$ BAE  $$$\cong$$∆$ ABD  ( By AAS rule  )
So,
AE  =  BD  ( By CPCT rule ) 

Regards