Question

In an isosceles triangle ABC , AB = AC and BD Perpendicular AC. Prove that (BD²-CD²) = 2CD × AD.

Answer 1

Hii friend,

Given: A ∆ABC in which

AB = AC and BD Perpendicular AC.

To prove : (BD²-CD²) = 2CD × AD

PROOF : From right ∆ADB , we have

AB² = AD²+BD² [ by pythagoras theroem ]

AC² = AD²+BD². [ AB = AC]

(CD+AD)² = AD²+BD² [ AC = CD + AD]

CD²+AD² + 2CD × AD = AD²+BD²

(BD²-CD²) = 2CD × AD

Hence,

(BD²-CD²)²= 2CD × AD ........proved.......


HOPE IT WILL HELP YOU...... :-)

Answer 2

Hey Friend ☺

Given :- In an isosceles triangle ABC ,

AB = AC

BD perpendicular AC

To Prove : - ( BD^2 - CD^2 ) = 2CD × AD

Proof :-

In triangle ABD ,

< BDA = 90

So by phythagorus theorem ,

( AB )^2 = ( BD )^2 + ( AD )^2

Since AB = AC , the above equation can be written as

( AC )^2 = ( BD )^2 + ( AD )^2

And also AC = AD + DC

》( AD + DC )^2 = ( BD )^2 + ( AD )^2

》( AD )^2 + ( DC )^2 + 2 ( AD )( CD ) = ( BD )^2+ ( AD )^2

》( CD )^2 + 2CD × AD = ( BD )^2

》( BD )^2 - ( CD )^2 = 2CD × AD

Hence it is proved .

Refer the attached figure

Hope it helps you ..!!

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