In the given figure, CD is perpendicular bisector of AB. EF is

Q) In the given figure, CD is perpendicular bisector of AB. EF is perpendicular to CD. AE intersects CD at G. Prove that CF/CD = FG/DG.

Ans:

Given that:

CD is perpendicular bisector of AB,

$\therefore$ AD = BD, ∠ CDB = ∠ GDA = 90⁰

EF is perpendicular bisector of CD,

$\therefore$ ∠ EFC = ∠ EFG = 90⁰

Let’s look at Δ CEF and Δ CBD:

∠ EFC = ∠ BDC = 90⁰(from given information)

∠ CEF = ∠ CBD (corresponding angles)

Therefore, Δ CEF $\sim$ Δ CBD (by AA similarity identity)

Hence, $\frac{CF}{CD} = \frac{EF}{BD}$

Since AD = BD, therefore,

Hence, $\frac{CF}{CD} = \frac{EF}{AD}$ …………. (i)

Let’s look at Δ EFG and Δ GDA:

∠ EFG = ∠ GDA = 90⁰(from given information)

∠ EGF = ∠ AGD (Interior angles)

Therefore, Δ EFG $\sim$ Δ GDA (by AA similarity identity)

Hence, $\frac{EF}{AD} = \frac{FG}{DG}$ ….. (ii)

From equation (i) and (ii), we get:

$\frac{CF}{CD} = \frac{FG}{DG}$

Hence Proved.

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