Question

Altitudes AD and BE of a triangle ABC are such that AE = BD. Prove that AD = BE.​

Answer 1

Answer:

Step-by-step explGiven  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:

Answer:

Step-by-step explGiven 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.