In the following figure, AE⊥BC, D is the mid point of BC, then x is equal to
In △ABC, AD is the median.
By Apollonius theorem,
AB2+AC2=2AD2+2DC2
c2+b2=2d2+2(12BC)2
c2+b2=2d2+a22
But, c2=h2+(a2−x)2
∴h2+(a2−x)2+b2=2d2+a22
h2+a24−ax+x2+b2=2d2+a22
But h2=d2−x2
d2−x2−ax+x2+b2=2d2+a24
ax=b2−d2−a24
x=1a[b2−d2−a24]
Hence proved.