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 \(\angle MAN = \angle MCN\).
The diagonals of the trapezium \(ABCD\) with the bases \(AD\) and \(BC\) intersect at the point \(O\); the points \(B'\) and \(C'\) are symmetrical to the vertices \(B\) and \(C\) with respect to the bisector of the angle \(BOC\). Prove that \(\angle C'AC = \angle B'DB\).
On the circle, the points \(A, B, C\) and \(D\) are given. The lines \(AB\) and \(CD\) intersect at the point \(M\). Prove that \(AC \times AD / AM = BC \times BD / BM\).
In the triangle \(ABC\), the height \(AH\) is drawn; \(O\) is the center of the circumscribed circle. Prove that \(\angle OAH = | \angle B - \angle C\)|.
Prove that from the point \(C\) lying outside of the circle we can draw exactly two tangents to the circle and the lengths of these tangents (that is, the distance from \(C\) to the points of tangency) are equal.
Two circles intersect at points \(A\) and \(B\). Point \(X\) lies on the line \(AB\), but not on the segment \(AB\). Prove that the lengths of all of the tangents drawn from \(X\) to the circles are equal.
Let \(a\) and \(b\) be the lengths of the sides of a right-angled triangle and \(c\) the length of its hypotenuse. Prove that:
a) The radius of the inscribed circle of the triangle is \((a + b - c)/2\);
b) The radius of the circle that is tangent to the hypotenuse and the extensions of the sides of the triangle, is equal to \((a + b + c)/2\).
Two circles touch at a point \(A\). A common (outer) tangent touching the circles at points \(C\) and \(B\) is drawn. Prove that \(\angle CAB = 90^{\circ}\).