Let \(AA_1\) and \(BB_1\) be the heights of the triangle \(ABC\). Prove that the triangles \(A_1B_1C\) and \(ABC\) are similar.
The bisector of the outer corner at the vertex \(C\) of the triangle \(ABC\) intersects the circumscribed circle at the point \(D\). Prove that \(AD = BD\).
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 a circle, the points \(A, B, C, D\) are given in the indicated order. \(M\) is the midpoint of the arc \(AB\). We denote the intersection points of the chords \(MC\) and \(MD\) with the chord \(AB\) by \(E\) and \(K\). Prove that \(KECD\) is an inscribed quadrilateral.
Two circles intersect at the points \(P\) and \(Q\). Through the point \(A\) of the first circle, the lines \(AP\) and \(AQ\) are drawn intersecting the second circle at points \(B\) and \(C\). Prove that the tangent at point \(A\) to the first circle is parallel to the line \(BC\).
The isosceles trapeziums \(ABCD\) and \(A_1B_1C_1D_1\) with corresponding parallel sides are inscribed in a circle. Prove that \(AC = A_1C_1\).
From the point \(M\), moving along a circle the perpendiculars \(MP\) and \(MQ\) are dropped onto the diameters \(AB\) and \(CD\). Prove that the length of the segment \(PQ\) does not depend on the position of the point \(M.\)
From an arbitrary point \(M\) on the side \(BC\) of the right angled triangle \(ABC\), the perpendicular \(MN\) is dropped onto the hypotenuse \(AB\). Prove that \(\angle MAN = \angle MCN\).