A trapezium encloses half circle bisecting its diagonals. Find angle of the trapezium. Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$ such that the circle $S$ with $AB$ as its diameter touches $CD$. Further, the circle $S$ passes through the midpoints of the diagonals $AC$ and $BD$ of the trapezium. The smallest angle of the trapezium is ______. Can anyone please tell me how to proceed? I think some more information needed to solve the problem.

The triangles AFO and ACB are similar since AO = OB, AF = FC and shared angle $\alpha$, , which leads to BC = 2OF = 2r. From the right triangle BEC, we have $\sin\angle$BCE = BE/BC = $\frac12$. Thus, the angle is 30$^\circ$.

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