Prove Angles are Equal

Question

Vertex $A$ of an acute triangle $ABC$ is connected by a segment with center $O$ of the circumscribed circle from vertex $A$ height $AH$ is drawn. Prove that $\angle BAH = \angle OAC$.

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So I don't get what the question is saying by "connected by a segment with center $O$ of the circumscribed circle". Is it saying that inside of the triangle, there's a circle and vertex $A$ is connected to it? And also, how would I prove this?

Answer 1

Hint: $\angle CDA=\angle CBA$ as they are angle that subtend equal arcs. Also, use the fact that $\triangle ACD$ and $\triangle AHB$ are right triangle.

Answer 2

Because $$\measuredangle OAC=\frac{1}{2}\left(180^{\circ}-\measuredangle AOC\right)=90^{\circ}-\measuredangle ABC=\measuredangle BAH.$$