Question

AD is the mid point of side BC in triangle ABC then prove that AB²+AC²=2(AD+BD)²

Answer 1

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Answer 2

$\huge\textbf\red{Answer :}$


Given:- AD is median. So BD = DC

Construction:- Draw AE⊥ BC

To prove:- AB2 + AC2 = 2 (AD2 + BD2)

Proof:- In ∆AED

By Pythagoras theorem

AD² = AE² + DE²

AE² = AD² - DE² ...... (1)

In ∆AED

By Pythagoras theorem

AD² = AE² + DE²

AE² = AD² - DE² ....... (2)


In ∆AEB

By Pythagoras theorem

AB² = AE² + BE²

AB² = AD² - DE² + BE² [From (2)]

Now BE = DE + BD

= (DE + BD)² + AD² - DE²

= DE² + BD² + 2DE. BD + AD² - DE²

AB² = BD² + AD² + 2DE. BD ....... (3)

In ∆AEC

AC² = AE² + EC²

= AD² - DE² + EC² [From (1)]

= AD² - DE² + (DC - DE)²

= AD² - DE² + DC² + DE² - 2DC. DE

= AD² + DC² - 2DC. DE

But BD = DC

So,

AC² = AD² + BD² - 2BD. DE ........ (4)

Add eq. (3) and (4)

AB² + AC² = BD² + AD² + AD² + BD² + 2DE. BD - 2BD. DE

AB² + AC² = BD² + AD² + AD² + BD² + 2BD. DE - 2BD. DE

AB² + AC² = 2AD² + 2BD²

AB² + AC² = 2(AD² + BD²)

Hence proved....

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