Problems

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The vertex \(A\) of the acute-angled triangle \(ABC\) is connected by a segment with the center \(O\) of the circumscribed circle. The height \(AH\) is drawn from the vertex \(A\). Prove that \(\angle BAH = \angle OAC\).

The vertex \(A\) of the acute-angled triangle \(ABC\) is connected by a segment with the center \(O\) of the circumscribed circle. The height \(AH\) is drawn from the vertex \(A\). Prove that \(\angle BAH = \angle OAC\).

From an arbitrary point \(M\) lying within a given angle with vertex \(A\), the perpendiculars \(MP\) and \(MQ\) are dropped to the sides of the angle. From point \(A\), the perpendicular \(AK\) is dropped to the segment \(PQ\). Prove that \(\angle PAK = \angle MAQ\).

On the circle, the points \(A, B, C\) and \(D\) are given. The lines \(AB\) and \(CD\) intersect at the point \(M\). Prove that \(AC \times AD / AM = BC \times BD / BM\).