[justify][size=150]If d[math]_i[/math], i = 1, 2, . . . , n, are the cevians with an end at the vertex of the right angle BÂC of the right triangle ABC and which section the hypotenuse BC in n + 1 congruent segments, then [math]\sum^n_{i=1}d_i^2=\frac{n\left(2n+1\right)}{6\left(n+1\right)}a^2[/math], where a is the measure of the hypotenuse of the right triangle. [/size][/justify]