Problems

Three circles are constructed on a triangle, with the medians of the triangle forming the diameters of the circles. It is known that each pair of circles intersects. Let \(C_{1}\) be the point of intersection, further from the vertex \(C\), of the circles constructed from the medians \(AM_{1}\) and \(BM_{2}\). Points \(A_{1}\) and \(B_{1}\) are defined similarly. Prove that the lines \(AA_{1}\), \(BB_{1}\) and \(CC_{1}\) intersect at the same point.

Point \(A\) is the centre of a circle and points \(B,C,D\) lie on that circle. The segment \(BD\) is a diameter of the circle. Show that \(\angle CAD = 2 \angle CBD\). This is a special case of the inscribed angle theorem.

Point \(A\) is the centre of a circle and points \(B,C,D\) lie on that circle. Show that \(\angle CAD = 2 \angle CBD\). This statement is known as the inscribed angle theorem and is used widely in Euclidean geometry.

Let \(BCDE\) be a quadrilateral inscribed in a circle with centre \(A\). Show that angles \(\angle CDE\) and \(\angle CBE\) are equal. Also show that angles \(\angle BCD\) and \(\angle BED\) are equal. This says that all angles at the circumference subtended by the same arc are equal.

Let \(BCDE\) be an inscribed quadrilateral. Show that \(\angle BCD + \angle BED = 180^{\circ}\).

The points \(B\),\(C\),\(D\),\(E\),\(F\) and \(G\) lie on a circle with centre \(A\). The angles \(\angle CBD\) and \(\angle EFG\) are equal. Prove that the segments \(CD\) and \(EG\) have equal lengths.

On the diagram below find the value of the angles \(\angle CFD\) and \(\angle CGD\) in terms of angles \(\angle CBD\) and \(\angle BDE\).

Point \(A\) is the centre of a circle. Points \(B,C,D,E\) lie on the circumference of this circle. Lines \(BC\) and \(DE\) cross at \(F\). We label the angles \(\angle BAD =\delta\) and \(\angle CAE = \gamma\). Express the angle \(DFB\) in terms of \(\gamma\) and \(\delta\).

On the diagram below \(BC\) is the tangent line to a circle with the centre \(A\), and it is known that the angle \(\angle ABC = 90^{\circ}\). Prove that the angles \(\angle DEB\) and \(\angle DBC\) are equal.

The triangle \(BCD\) is inscribed in a circle with the centre \(A\). The point \(E\) is chosen as the midpoint of the arc \(CD\) which does not contain \(B\), the point \(F\) is the centre of the circle inscribed into \(BCD\). Prove that \(EC = EF = ED\).