In triangle ABC, AB = 10 cm. Through point K on the side AB, a straight line KM is drawn parallel

In triangle ABC, AB = 10 cm. Through point K on the side AB, a straight line KM is drawn parallel to AC, AK = 5 cm. prove that BM = MC.

The ABC triangle is similar to the CME triangle in two angles. The angle B in triangles is common, the angle BAC = BKM as the corresponding angles at the intersection of parallel straight lines KM and AC secant AB.

AB = 10 cm, AK = 5 cm, then BK = AB – AK = 5 cm. The coefficient of similarity of triangles is: AB / KB = 10/5 = 2.

Then CB / BM = 2. CB = 2 * BM, since CB = BM + CM, then

BM + CM = 2 * BM.

CM = BM, as required.