In the given figure, $CD$ is the perpendicular bisector of $AB$. $EF$ is perpendicular to $CD$. $AE$ intersects $CD$ at $G$. Prove that $\frac{CF}{CD} = \frac{FG}{DG}$.

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Step 1: $	riangle EFG \sim 	riangle ADG$ (AA similarity)
$\Rightarrow \frac{EF}{AD} = \frac{FG}{DG}$ ... (i)
Step 2: $	riangle EFC \sim 	riangle BDC$ (AA similarity)
$\Rightarrow \frac{EF}{BD} = \frac{CF}{CD}$
Step 3: $\Rightarrow \frac{EF}{AD} = \frac{CF}{CD}$ {since $BD = AD$} ... (ii)
Step 4: Using (i) and (ii): $\frac{FG}{DG} = \frac{CF}{CD}$

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