Question

plzzz... solve the sum

Answer 1

Hope it helps u pls marlk as brainliest.

Answer 2

Hey !!

Here is your answer...


Given :- In∆ABC, AB = AC and d is a point on BC.

To Prove :- AB^2 - AD^2 = BD × CD

Proof :- AE perpendicular to BC.

In ∆ABE and ∆ACE,
AB = AC. ...... ( given )
AE = AE. .......( common )
angle AEB = angle AEC. ..... ( right angle )

∆ABE ~ ∆ACE. .... ( by RHS theorem )

BE = CE

In right angled triangle ∆ABE,
AB^2 = AE^2 + BE^2. .....(1)


In right angled triangle∆ADE,
AD^2 = AE^2 + DE^2. ..... (2)

Now, Subtract eq.(2) form eq. (1)

AB^2 - AD^2 = ( AE^2 + BE^2 ) - ( AE^2 + DE^2 )
AB^2 - AD^2 = AE^2 + BE^2 - AE^2 - DE^2
AB^2 - AD^2 = BE^2 - DE^2
AB^2 - AD^2 = ( BE - DE ) ( BE + DE )

As we know BE = CE

AB^2 - AD^2 = ( BE - DE ) ( CE + DE )
AB^2 - AD^2 = BD × CD. .... ( proved )


HOPE IT HELPS YOU.

THANKS
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