In a trapezoid with bases AD and BC, the diagonals intersect at point O, BO: OD = 3: 2, AC = 25 cm. Find OC and AO.

Let us prove that triangle BOS is similar to triangle AOD.

Angle BOC = AOD as vertical angles at the intersection of diagonals AC and BD.

Angle ВСО = CAD as criss-crossing angles at the intersection of parallel straight lines ВС and АD of secant АС.

Then the triangles BOC and AOD are similar in two angles.

In similar triangles BOС and AOD, by condition, BO / OD = 3/2, then BO / OD = OC / OA = 3/2.

Then 3 * OA = 2 * OC.

AC = 25 cm, then OA = 25 – OC.

3 * (25 – OС) = 2 * OS.

75 – 3 * OС = 2 * OS.

OS = 75/5 = 15 cm, then OA = 25 – 15 = 10 cm.

Answer: The length of the OA segment is 10 cm, the length of the OS is 15 cm.