Problems

Point $A$ is the centre of a circle and points $B,C,D$ lie on that circle. Show that $\mathrm{\angle }CAD=2\mathrm{\angle }CBD$. 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 $\mathrm{\angle }CDE$ and $\mathrm{\angle }CBE$ are equal. Also show that angles $\mathrm{\angle }BCD$ and $\mathrm{\angle }BED$ 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 $\mathrm{\angle }BCD+\mathrm{\angle }BED={180}^{\circ }$.

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

On the diagram below find the value of the angles $\mathrm{\angle }CFD$ and $\mathrm{\angle }CGD$ in terms of angles $\mathrm{\angle }CBD$ and $\mathrm{\angle }BDE$.

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 $\mathrm{\angle }BAD=\delta$ and $\mathrm{\angle }CAE=\gamma$. 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 $\mathrm{\angle }BAD$ and $\mathrm{\angle }CAE$ 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 $\mathrm{\angle }ABC={90}^{\circ }$. Prove that the angles $\mathrm{\angle }DEB$ and $\mathrm{\angle }DBC$ 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 $EC=EF=ED$.

Picasso got a new set of crayons. He started colouring various things. First, Picasso coloured a line under the following condition: each point on a line is coloured either red or blue. Show that there are three different points $A,B,C$ on the line of the same colour such that $AB=BC$.