Problems

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.

Let \(ABC\) be a triangle with given angles \(\angle BAC\) and \(\angle ABC\). What is the value of the angle \(\angle BCD\) in terms of \(\angle BAC\) and \(\angle ABC\)?

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 this circle. Lines \(BC\) and \(DE\) cross at \(F\). Angles \(\angle BAD =\delta\), \(\angle CAE = \gamma\). Express the angle \(DFB\) in terms of \(\gamma\) and \(\delta\).

The triangle \(ABC\) is inscribed into the circle with centre \(E\), the line \(AD\) is perpendicular to \(BC\). Prove that the angles \(\angle BAD\) and \(\angle CAE\) are equal.

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.