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Asked by 21st February 2009, 4:02 PM
Answered by Expert
Answer:

Given: A triangle ABC in which AD is the median
bisecting BC.
To prove that: AB^2+AC^2=2(BD^2+AD^2)
Here there could be 2 possibilities,
1. Both the base angles are acute or
2. One of the base angles is obtuse.
Contruct:Draw AE perpendicular to BC
( or BC produced if necessary)


Proof: If the 2 angles are unequal, then one of them will be
acute and another obtuse.
Let In triangle ADB,
AB^2=AD^2+BD^2+2*BD*DE      ...(1)
And in triangle ADC
AC^2=AD^2+DC^2-2*CD*DE      ...(2)
But since CD=BD,
Therefore
CD^2=BD^2 and BD*DE=CD*DE
By adding (1) & (2), we get
AB^2+AC^2=2(BD^2+AD^2)

Answered by Expert 22nd February 2009, 1:39 PM
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