Question

In triangle ABC. AD is drawn perpendicular to BC. Prove that AB²-BD²= AC² - CD²

Answer 1

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In ∆ ABD, /_ADB = 90°

By applying Pythagoras Theorem

( H )² = ( P )² + ( B )²

( AB )² = ( AD )² + ( BD )²

AB² = AD² + BD²

AD² = AB² - BD² ___________( 1 )

Again,

In ∆ ACD, /_ADC = 90°

By Applying Pythagoras Theorem,

( H )² = ( P )² + ( B )²

( AC )² = ( AD )² + ( CD )²

AC² = AD² + CD²

AD² = AC² - CD² __________( 2 )

From ( 1 ) and ( 2 ), we get

AB² - BD² = AC² - CD²

Hence, proved.

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Answer 2

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