[size=150][justify]If d[math]_i[/math], i = 1, 2, . . . , n, are the cevians that section an internal angle of a triangle at n + 1 congruent angles, then [math]d_i^2=bc\frac{d_i}{d_{n+1-i}}-\left(k_1+k_2+...+k_i\right)\left(k_{i+1}+...+k_{n+1}\right)[/math], where b and c are the measurements of the sides that determine the sectioned angle and k[math]_1[/math], k[math]_2[/math], . . ., k[math]_{n+1}[/math] are the measurements of the segments determined by the cevians on the side opposite to the sectioned angle.[/justify][/size]