The area of the ADF triangle is 16 cm2. Find the area of the triangle ABC if DF is parallel to BC and AB = 9 cm, AD = 3 cm.

Let us prove that triangles ABC and ADF are similar.

Since, according to the condition, BC is parallel to DF, then the angle ABC = ADF is as criss-crossing angles at the intersection of parallel lines BC and DF of the secant BD, the angle A is common for triangles, then the triangles are similar in two angles.

The coefficient of similarity of triangles is: K = AD / AB = 3/9 = 1/3.

The areas of similar triangles are referred to as the squared coefficient of similarity.

Sadf / Savs = 1/9.

Sас = 9 * Sаdf = 9 * 16 = 144 cm2.

Answer: The area of the triangle ABC is 144 cm2.