Question

Two circles intersect each other at A and B.Let DC be a common tangent touching the circle in point C and D
Prove that $\angle \mathrm{CAD}+\angle \mathrm{CBD}=180°$

Answer 1

$$\begin{gathered} \angle \mathrm{BCD}=\angle \mathrm{CAB}...\left(1\right)(\mathrm{Alternate}\mathrm{segment}\mathrm{theorem}) \\ \angle \mathrm{CDB}=\angle \mathrm{DAB}...\left(2\right)(\mathrm{Alternate}\mathrm{segment}\mathrm{theorem}) \\ \mathrm{In}△\mathrm{BCD},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BCD}+\angle \mathrm{CDB}+\angle \mathrm{CBD}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \mathrm{Using}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}: \\ \angle \mathrm{CAB}+\angle \mathrm{DAB}+\angle \mathrm{CBD}=180° \\ \Rightarrow \angle \mathrm{CAD}+\angle \mathrm{CBD}=180°(\because \angle \mathrm{CAD}=\angle \mathrm{CAB}+\angle \mathrm{DAB}) \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$