**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,

AD = BD, ∠ CDB = ∠ GDA = 90^{0}

EF is perpendicular bisector of CD,

∠ EFC = ∠ EFG = 90^{0}

Let’s look at Δ CEF and Δ CBD:

∠ EFC = ∠ BDC = 90^{0 }(from given information)

∠ CEF = ∠ CBD (corresponding angles)

Therefore, Δ CEF Δ CBD (by AA similarity identity)

Hence,

Since AD = BD, therefore,

Hence, …………. (i)

Let’s look at Δ EFG and Δ GDA:

∠ EFG = ∠ GDA = 90^{0 }(from given information)

∠ EGF = ∠ AGD (Interior angles)

Therefore, Δ EFG Δ GDA (by AA similarity identity)

Hence, ….. (ii)

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

**Hence Proved.**