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

In the given figure, seg EF is a diameter and seg DF is a tangent segment.

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

In right ∆DEF,

DE² = EF² + DF² .....(1)

Using tangent secant segments theorem, we have

DE × DG = DF² .....(2)

Subtracting (2) from (1), we get

DE² − DE × DG = EF² + DF² − DF²

⇒ DE × (DE − DG) = EF²

⇒ DE × GE = (2r)² = 4r² (EF = 2r)

Hence, DE × GE = 4r²