Question

046. 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.​

Answer 1

Step-by-step explanation:

Here

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

∠ 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  )

therefore , ∆ BAE  ≅∆ ABD  (  AAS rule  )

So,

AE  =  BD  (  CPCT  )  

hence proved