In trapezoid ABCD with bases AD and BC, the diagonals intersect at point O, BC: AD = 3: 5, BD = 24 see Find BO and OD.

Consider two triangles BOC and AOD. In triangles, the angle BOC is equal to the angle AOD as vertical angles, the angle OBC is equal to the angle ODA, as criss-crossing angles at the intersection of parallel straight lines AD and BC secant BD. Then the triangles BOC and AOD are similar in the first sign of similarity, in two angles.

By the condition ВС / АD = 3/5, therefore, ВС / ОD = 3/5.

Segment BO = ВD – OD = 24 – OD.

Let’s substitute ВO in proportion.

(24 – OD) / OD = 3/5.

5 * (24 – OD) = 3 * OD.

8 * OD = 120.

OD = 10/8 = 15 cm.

Then ВO = 24 – 15 = 9 cm.

Answer: ВO = 9 cm, OD = 15 cm.