Question

Prove that in any triangle, the ratio of sum of squares of the sides to the sum of the squares of its median is 4:3

Answer 1

here's the solution of the problem in the picture

Answer 2

Step-by-step explanation:

By apollonius theorem

AB2 + AC2 = 2BD2 + 2AD2AB2 + AC2 = 2{left( {frac{{BC}}{2}} right)^2} + 2A{D^2}

AB2 + AC2 = frac{{B{C^2}}}{2} + 2A{D^2}

2AB2 + 2AC2 = BC2 + 4AD2 .....(1)

Similarly 2AB2 + 2BC2 = AC2 + 4BE2 ....(2)

and 2BC2 + 2AC2 = AB2 + 4CF2 ......(3)

Adding (1), (2) and (3)

4AB2 + 4BC2 + 4AC2 = AB2 + BC2 + AC2 + 4AD2 + 4BE2 + 4CF2

3 (AB2 + BC2 + AC2) = 4(AD2 + BE2 + CF2)

frac{{A{B^2} + B{C^2} + A{C^2}}}{{A{D^2} + B{E^2} + C{F^2}}} = frac{4}{3}

frac{{{text {Sum of squares of the sides}}}}{{{text {sum of squares of its median}}}} = frac{4}{3}