Answer :

Given: A ΔABC right-angled at B, and D is the mid-point of BC, i.e. BD = CD

To Prove: AC^{2} = (4AD^{2} - 3AB^{2})

Proof:

In ΔABD,

By Pythagoras theorem, [i.e. Hypotenuse^{2} = Base^{2}+ Perpendicular^{2}]

AD^{2} = AB^{2} + BD^{2}

[ as D is mid-point of BC, therefore,

⇒ 4AD^{2} = 4AB^{2} + BC^{2}

⇒ BC^{2} = 4AD^{2} - 4AB^{2} [1]

Now, In ΔABC, again By Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

AC^{2} = AB^{2} + 4AD^{2} - 4AB^{2} [From 1]

AC^{2} = 4AD^{2} - 3AB^{2}

Hence Proved !

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