In triangle ABC AB = 15 cm, AC = 20 cm, BC = 32 cm. On the AB side, point D is taken, and on the AC

In triangle ABC AB = 15 cm, AC = 20 cm, BC = 32 cm. On the AB side, point D is taken, and on the AC side, point E, so that DE is parallel to BC. Find the length DE if BD = 6 cm, CE = 8 cm.

Let us prove that triangles ABC and ADE are similar.

The angle A of the triangles is common. By condition, DE is parallel to BC, then the angle AED = ACB as the corresponding angles at the intersection of parallel straight lines DE and BC secant AC.

Then triangles ABC and ADE are similar in two angles.

The length of the segment AE = AC – CE = 20 – 8 = 12 cm.

Then the coefficient of similarity of triangles is: K = AE / AC = 12/20 = 3/5.

DE / ВС = 3/5.

DE = 3 * BC / 5 = 3 * 32/5 = 19.2 cm.

Answer: The length of the segment DE is 19.2 cm.