Question

Suppose ABC is a triangle inscribed in a circle , the bisector of$\angle$ABC intersects the circle again in D, the tangent at D intersect the line BA and line BC in E and F respectively
Prove that $\angle \mathrm{EDA}=\angle \mathrm{FDC}$

Answer 1

$$\begin{gathered} \angle \mathrm{ADE}=\angle \mathrm{ABD}...\left(1\right)(\mathrm{Angles}\mathrm{in}\mathrm{alternate}\mathrm{segments}) \\ \angle \mathrm{CDF}=\angle \mathrm{CBD}...\left(2\right)(\mathrm{Angles}\mathrm{in}\mathrm{alternate}\mathrm{segments}) \\ \mathrm{Using}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{ADE}=\angle \mathrm{CDF} \end{gathered}$$