In trapezoid ABCD with bases AD and BC, the diagonals intersect at point O. BO = 4 cm,

In trapezoid ABCD with bases AD and BC, the diagonals intersect at point O. BO = 4 cm, OD = 8 cm, AC = 15 cm. Find the lengths of OС and AO.

Let us prove that the triangles BOC and AOD are similar.

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

Angle OBC = ODA as criss-crossing angles at the intersection of parallel straight lines AD and BC of the secant ВD.

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

Let the segment OC = X cm, then OA = (15 – X) cm.

From the similarity of triangles:

ОВ / ОD = ОВ / ОА.

4/8 = X / (15 – X).

15 – X = 2 * X.

3 * X = 15.

X = OC = 15/3 = 5 cm.

ОА = 15 – 5 = 10 cm.

Answer: The length of the segment OС = 5 cm, OA = 10 cm.