In the given figure, DEFG is a square. $\displaystyle \triangle \mathrm{ABC}$ is right angle triangle with $\displaystyle \angle \mathrm{A}=90^{\circ}$. Prove that $\displaystyle \mathrm{AG} \times \mathrm{DG}=\mathrm{AF} \times \mathrm{DB}$.

DEFG is a square. \(\displaystyle \therefore \mathrm{GF} \| \mathrm{BE}\) \[\Longrightarrow \angle \mathrm{GBD}=\angle \mathrm{AGF} \] \[\therefore \triangle \mathrm{GAF} \sim \triangle \mathrm{BDG} \] \[\Rightarrow \frac{\mathrm{AG}}{\mathrm{DB}}=\frac{\mathrm{AF}}{\mathrm{DG}} \] \[\Rightarrow \mathrm{AG} \times \mathrm{DG}=\mathrm{AF} \times \mathrm{DB} \]