An acute-angled triangle ABC is given. Let AD, CE, BM be its heights, CD = DE = 7, DM = 8. Find CB.

1) Triangle BCE is similar to triangle CAE (in 2 corners), angle A is common.
2) In the triangle ABC, ED is the middle line, which means ED || AC, AC = ED * 2, thus AC = 7 * 2 = 14 cm.
3) In triangle BCA, DM is the middle line, so DM || AB, AB / 2 = DM, thus AB = 2DM = 6 * 2 = 12 cm.
4) The BDM triangle is rectangular, so we use the Pythagorean theorem:
(BM) ^ 2 = (BD) ^ 2 + (DM) ^ 2;
(BM) ^ 2 = 10 ^ 2 + 6 ^ 2;
(BM) ^ 2 = 136;
BM = 2√34.
5) The IUD triangle is also rectangular, so we use the Pythagorean theorem:
(10x) ^ 2 = BM ^ 2 + MC ^ 2.
AC in triangle ABC is halved by height and median BM.
100x ^ 2 = (2√34) ^ 2 + 7 ^ 2;
100x ^ 2 = 4 * 34 + 49;
100x ^ 2 = 136 + 49;
100x ^ 2 = 185;
x ^ 2 = 185/100;
x = √1.85
BC = 10√1.8.