In the given figure, CD is the perpendicular bisector of AB . EF is perpendicular to CD . AE intersects CD at G .…

CBSE Class 10 Maths PYQ · Triangles · Similarity with Triangles · 3 Marks · March 2023 · Standard

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953 Marks · March 2023 · Standard

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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$\triangle EFG \sim \triangle ADG$
$\Rightarrow \frac{EF}{AD} = \frac{FG}{DG}$
quad (i)
$\triangle EFC \sim \triangle BDC$
$\Rightarrow \frac{EF}{BD} = \frac{CF}{CD}$
$\Rightarrow \frac{EF}{AD} = \frac{CF}{CD}$
quad $\{BD = AD\}$
quad (ii)
Using (i) and (ii)
$\frac{FG}{DG} = \frac{CF}{CD}$