In a convex rectangle ABCD, AB = 9 cm, BC = 8 cm, CD = 16 cm AD = 6 cm, BD = 12 cm Prove that ABCD is a trapezoid.

Consider triangles ABD and BCD and check their similarity.

BD / CD = 12/16 = 0.75.

AD / BC = 6/8 = 0.75.

AB / BD = 9/12 = 0.75.

Since the ratio of the respective sides is equal, the triangles are similar in the third sign of similarity of triangles.

Since triangles ABD and BCD are similar, the angle ABD = CDB.

These angles in the quadrangle ABCD are crosswise at the intersection of straight lines AB and CD of the secant BD. But since the angle ABD = CDB, the straight lines AB and CD are parallel, and therefore ABCD is a trapezoid with bases AB and CD, which was required to be proved.