In right-angled triangle $ A B C $ in which $ \angle C=90 $, if $ D $ is the mid-point of $ B C $ prove that $ A B^{2}=4 A D^{2}-3 A C^{2} $.

Given:

 In right-angled triangle \( A B C \), \( \angle C=90 \) and \( D \) is the mid-point of \( B C \).

To do:

We have to prove that \( A B^{2}=4 A D^{2}-3 A C^{2} \).

Solution:

\( D \) is the mid-point of \( B C \). Therefore, $BC=2CD=2BD$.

In $\triangle ABC$, by Pythagoras theorem,

$AB^2=AC^2+BC^2$

$AB^2=AC^2+(2CD)^2$    ($BC=2CD$)

$AB^2=AC^2+4CD^2$.....(i)

In $\triangle ACD$, by Pythagoras theorem,

$AD^2=AC^2+CD^2$

$CD^2=AD^2-AC^2$....(ii)

Substituting equation (ii) in equation (i), we get,

$AB^2=AC^2+4(AD^2-AC^2)$

$AB^2=AC^2+4AD^2-4AC^2$

$AB^2=4AD^2-3AC^2$

Hence proved.

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Updated on: 10-Oct-2022

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