Question

In the adjoining 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: seg EF is the diameter. seg DF is a tangent to the circle, radius = r

To prove: DE × GE = 4r Construction: Join seg GF. 

Proof: 

seg EF is the diameter. [Given] 

∴ ∠EGF = 90° (i) [Angle inscribed in a semicircle] seg DF is a tangent to the circle at F. [Given]

∴ ∠EFD = 90° (ii) [Tangent theorem] In ∆DFE, 

∠EFD = 90 ° [From (ii)] 

seg FG ⊥ side DE [From (i)] 

∴ ∆EFD ~ ∆EGF [Similarity of right angled triangles] 

∴ (EF/GE) = (DE/EF) [Corresponding sides of similar triangles] 

∴ DE × GE = EF² 

∴ DE × GE = (2r)² [diameter = 2r] 

∴ DE × GE = 4r²