Prove that the sum of the squares of the medians of a right-angled triangle is one and a half times the square of the hypotenuse.

Let the triangle ABC be given, AB = c – hypotenuse, BC = a and AC = b – legs, m1; m2; m3 is the median. Let’s prove that m1 ^ 2 + m2 ^ 2 + m3 ^ 2 = 1.5 * c ^ 2.

For the proof we use right-angled triangles CBM2; ACM1; and ABC, where M1; M2; M3 – middle of the aircraft; AC; AB,

1) From triangle АСМ1: в ^ 2/4 + a ^ 2 = m2 ^ 2;

2) From the triangle BCM2: b ^ 2 + a ^ 2/4 = m1 ^ 2;

3) m3 = s / 2; m3 ^ 2 = c ^ 2/4; add 1) + 2) + 3), we get:

4) b ^ 2/4 + a ^ 2 + b ^ 2 + a ^ 2/4 + c ^ 2/4 = m2 ^ 2 + m1 ^ 2 + m3 ^ 2;

5) 5/4 * (b ^ 2 + a ^ 2) + c ^ 2/4 = m2 ^ 2 + m1 ^ 2 + m3 ^ 2; 6/4 * c ^ 2 = 1.5 * c ^ 2 = m2 ^ 2 + m1 ^ 2 + m3 ^ 2.