A property about harmonic quadrilateral

Question

Point $A$ is the center of the circle. $BA\bot BE, FA\bot FE$. Prove $\displaystyle \frac{CG}{GD}=\frac{CE}{ED}$.

Answer 1

Since $\triangle BGD$ is similar to $\triangle CGF$, $$CG:BG=CF:BD\tag{1}.$$ Similarly $\triangle BGC$ is similar to $\triangle DGF$, so $$BG:DG=BC:DF\tag{2}.$$ From (1) and (2), $$\frac{CG}{DG}=\frac{CG}{BG}\cdot\frac{BG}{DG}=\frac{CF}{BD}\cdot\frac{BC}{DF}=\frac{CF}{DF}\cdot\frac{BC}{BD}\tag{3}.$$ On the other hand, $\triangle BED$ is similar to $\triangle CEB$, so $$BC:BD = CE:BE\tag{4},$$ and $\triangle CFE$ is similar to $\triangle FDE$, so $$CF:DF = EF:DE\tag{5}.$$ From (3), (4), and (5), noting that $BE=EF$, we have $$\frac{CG}{DG}=\frac{EF}{DE}\cdot\frac{CE}{BE}=\frac{CE}{DE}.$$