Question

solve this question.....

Answer 1

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Given :

ΔABC with a median AD

∴ BD = CD

To Prove :

$AB^2 + AC^2 = 2( AD^2 + BD^2)$

$AB^2 + AC^2 = 2AD^2 + 2 BD^2$

Construction :

Draw AE ⊥ BC to meet BC at E

Proof :

In ΔAEB

$AB^2 = AE^2 + BE^2$

$AB^2 = AE^2 + (BD - ED )^2$

$AB^2 = AE^2 + BD^2 +ED^2 - 2BD * ED$ ------> 1

In ΔAEC

$AC^2 = AE^2 + CE^2$

$AC^2 = AE^2 + (CD + ED)^2$

$AC^2 = AE^2 + CD^2 + ED^2 + 2CD * ED$

$AC^2 = AE^2 + CD^2 + ED^2 + 2BD* ED$--------> 2

Adding 1 and 2

$AB^2 + AC^2 = 2AE^2 + 2BD^2 + 2ED^2$

⇒ $2 ( AE^2 + ED^2 + BD^2 )$

⇒ $2 ( AD^2 + BD^2 )$

Hence Proved

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Answer 2

Answer:

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Step-by-step explanation:

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