Question

In the given figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that, DE × GE = 4r²

Answer 1

Given:

In the given figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r.

To prove, DE × GE = 4r²

seg DF ⊥ seg HF

∴ ∠HFD = 90º   ...(Tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Using the tangent secant segments theorem, we have

DF² = DE × DG    ...(1)

Now, In ΔDFE, ∠F = 90º

By Pythagoras’ theorem,

DE² = DF² + EF²

DE² = DE × DG + (2r)²   ...(From eqn (1) and diameter = 2 × radius)

DE² = DE × DG + 4r²

DE² − DE × DG = 4r²

DE × (DE − DG) = 4r²

DE × GE = 4r²   ...(D - G - E)

Hence, DE × GE = 4r²