In trapezoid ABCD (AD and BC are bases) point K lies on the side of CD and CK: KD

In trapezoid ABCD (AD and BC are bases) point K lies on the side of CD and CK: KD = 1: 2. AK meets BD at point O. Prove that if BC: AD = 1: 2, then BO = OD.

Let us extend the segment AK to the intersection with the base of the BC. The KCH triangle is similar to the KAD triangle in two angles, Then:

СK / KD = CH / AD.

1/2 = CH / AD.

AD = 2 * CH.

Since BC / AD is also equal to 1/2, then AD = 2 * BC.

Then CH = BC = AD / 2, and AD = BH.

The triangles BOH and AOD are equal along the side and the angles adjacent to it.

BH = AD, angle HBO = ADO, angle BHO = DAO.

Then BO = OD, OA = HO, as required.