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Q ) In Triangle ABC , angle BAC =90 and AD is perpendicular to BC. Prove that AD^2= BD×DC


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Answer 1

$\: \: \: \: \: \: \: \bold{In \: \: \: \triangle \: ADC , }$


$\: \: \: \: \: \: \: \: \: \: \underline{ By \: \: Pythagoras \: \: Theorem , }$

= > AC² = DC² + AD²


= > AD² = AC² - DC² ------: ( 1 )





$\: \: \: \: \: \: \: \bold{In \: \: \: \triangle \: ABD, }$


$\: \: \: \: \: \: \: \: \: \: \: \underline{ By \: \: Pythagoras \: \: Theorem , }$


= > AB² = BD² + AD²


= > AD² = AB² - BD² -------: ( 2 )





$\: \: \: \: \: \: \: \: \bold{In \: \: \: \triangle \: ABC, }$


BC = BD + DC ----: ( 3 )



$\: \: \: \: \: \: \: \underline{By \: \: Pythagoras \: \: Theorem , }$



= > BC² = AC² + AB²



Putting the value of BC from ( 3 ) ,



= > ( BD + DC )² = AC² + AB²



By Formula,
( a + b )² = a² + b² + 2ab



= > BD² + DC² + 2{ BC × DC } = AC² + AB²



= > 2{ BD × DC } = AC² + AB² - BD² - DC²



= > 2{ BD × DC } = AC² - DC² + AB² - BD²




Putting the value(s) of [ AC² - DC² ] and [ AB² - BD² ] from ( 1 ) and ( 2 ) respectively,




= > 2{ BD × DC } = AD² + AD²


= > 2{ BD × DC } = 2 AD²


= > BD × DC = AD²





Hence, prove that BD × DC = AD².

Answer 2

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