# If in triangle ABC, AD is a median and AM is perpendicular to BC then prove that AB2= AD2 – BC × DM + 1/4 BC2.

Given: D is the mid-point of BC.

And AM BC.

To Prove:

Proof: In right-angled ∆ABM,

By Pythagoras theorem, we can write as

AB2 = AM2 + BM2

[ (hypotenuse)2 = (perpendicular)2 + (base)2] …(i)

In right-angled ∆AMD,

By Pythagoras theorem, we can write as

[ (hypotenuse)2 = (perpendicular)2 + (base)2] …(ii)

Subtract equation (ii) from equation (i), we get

AB2 – AD2 = AM2 – AM2 + BM2 – MD2

AB2 – AD2 = 0 + BM2 – MD2

AB2 = AD2 – MD2 + BM2

AB2 = AD2 – DM2 + (BD – DM)2

AB2 = AD2 – DM2 + BD2 + DM2 – 2(BD)(DM)

AB2 = AD2 + BD2 – 2(BD)(DM)

[ ]

Hence, proved.

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