Question

In a right triangle ABC,∆A=90°.If AD is drawn perpendicular to BC,then prove that AB²+CD²=BD²+AC²

Answer 1

Appropriate Question:

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• In a right triangle ∆ABC, ∠A = 90°.If AD is drawn perpendicular to BC,then prove that AB² + CD² = BD² + AC².

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Given: ABC is a right triangle and ∠A is 90°.

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Need to Prove: AB² + CD² = BD² + AC².

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As we given with a right triangle ABC in which ∠A is 90°.

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We need to prove that, AB² + CD² = BD² + AC².

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Let us draw AD ⊥ BC.

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Now, we have two right angle triangles, that is, ∆ABD and ∆ACD respectively.

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As we know that, if any triangle is a right angle triangle then that right angle triangle follow pythagoras theorem.

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In right angle ∆ABD, and using pythagoras theorem in it, we get:

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$\sf{:\implies AB^2= AD^2 + BD^2}$

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$\sf{:\implies AD^2 = AB^2 - BD^2 \quad -----(\bf{1})}$

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Similarly, in right angle ∆ACD and using pythagoras theorem in it, we get:

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$\sf{:\implies AC^2 = AD^2 + CD^2}$

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$\sf{:\implies AD^2 = AC^2 - CD^2 \quad -----(\bf{2})}$

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Now, from equation (1) and equation equation (2), we get:

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$\sf{:\implies AB^2 - BD^2 = AC^2 - CD^2}\\\\\\ \sf{:\implies AB^2 - BD^2 + CD^2 = AC^2}\\\\\\ \sf{:\implies AB^2 + CD^2 - BD^2 = AC^2} \\\\\\ \sf{:\implies \boxed{\rm{\pink{AB^2 + CD^2 = AC^2 + BD^2}}}}$

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⠀⠀⠀⠀⠀Hence proved!

Answer 2

Step-by-step explanation:

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.