Question

in an isosceles∆ABC, with AB=AC, BD perpendicular AC . Prove that:BD square -CD square =2CD×AD

Answer 1

Make it as brainliest!

Answer 2

$\underline { \blue { Given: }}$

$In \: an\: isosceles \triangle ABC, \:with \:AB=AC,\: BD\\ perpendicular \:to \:AC$

$\underline { \red { To \:prove }}$

$\pink { BD^{2} - CD^{2} = 2CD \times AD }$

$\underline { \green { Proof: }}$

$In \: \triangle ABD , \angle {ADB} = 90\degree$

$\blue {( By \: Pythagoras \:Theorem )}$

$AB^{2} = AD^{2} + BD^{2}$

$\implies AC^{2} = AD^{2} + BD^{2} \\ \blue { (given\: AB = AD ) }$

$\implies ( AD + DC )^{2} = AD^{2} + BD^{2}$

$\implies AD^{2} + DC ^{2} + 2 AD \times DC = AD^{2} + BD^{2}$

$\implies 2 AD \times DC = BD^{2} - DC^{2}$

$\therefore \green { BD^{2} - DC^{2} = 2 AD \times CD}$

$Hence\: proved$

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