In trapezoid ABCD with bases AD and BC, the diagonals meet at point O. Prove the similarity of triangles AOD and COB.

In triangles COB and AOD, the angle BOC and AOD are equal as the vertical angles at the intersection of the diagonals BD and AC.

The angle ВСО is equal to the angle ОАD as criss-crossing angles at the intersection of parallel straight lines АD and ВС secant АС.

Then the COB triangle is similar to the AOD triangle by the first criterion of similarity of triangles, in two angles, which was required to be proved.