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.

CD is perpendicular bisector 10th board CBSE

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.

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top