Given an isosceles trapezoid AB parallel to CD DO = 15 cm, OB = 9 cm, DC = 25 cm. Prove: AO: OC = BO: OD find AB

Let us prove that triangles AOB and COD are similar.

Angle AOB = COD as vertical angles at the intersection of diagonals.

AB parallel to CD, then the angle ABO = CDO as criss-crossing angles at the intersection of parallel lines AB and CD of the secant BD.

Then triangles AOB and COD are similar in two angles.

From the similarity of triangles, AO / OC = BO / OD, as required.

Determine the coefficient of similarity of triangles. K = OB / OD = 9/15 = 3/5.

Then AB / CD = 3/5.

AB = 3 * CD / 5 = 3 * 25/5 = 15 cm.

Answer: The length of the smaller base is 15 cm.