In trapezium ABCD, diagonals AC and BD are drawn, which meet at point O. Prove that triangle COB is similar to triangle AOD.
Since ABCD is a trapezoid, its bases are parallel, АD || BC.
In the triangles COB and AOD, the angle OBC = ODA as criss-crossing angles at the intersection of parallel straight lines ВС and АD of the secant ВD.
Angle BOC = AOD as vertical angles at the intersection of diagonals AC and BD.
Then the triangles COB and AOD are similar in two angles, as required.
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