Question

AD and BE are respectively the altitudes of triangle ABC with AB = AC. Prove that AE = BD

Answer 1

Given  AD and BE are altitude and AC =  BC  , S
Here

∠ BEA  =  ∠ BEC  = 90°  --------------- ( 1 )

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

∠ BEA  =  ∠ ADB  = 90°  ---------------- ( 3 )

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

∠ CAB  =  ∠ CBA  ----------------- ( 4 )

Now In ∆ BAE  and ∆ ABD 

∠ BEA  =  ∠ ADB  ( From equation 3 )

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

Hence

∆ BAE  ≅∆ ABD  ( By AAS rule  )
So,
AE  =  BD  ( By CPCT rule )  ( Hence proved )

Answer 2

Answer:

your answer is in the picture captured

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