Q. 44.6( 7 Votes )
If in triangle ABC, AD is a median and AM is perpendicular to BC then prove that AB2= AD2 – BC × DM + 1/4 BC2.
Answer :
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
AD2 = AM2 + MD2
[∵ (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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