In figure 3.81, 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: \(\triangle EFD\) is right angled at F and FG ⊥ ED, so \(EF^2 = EG \times ED\), i.e. \((2r)^2 = DE \times GE\).

Draw seg FG.

• \(\angle EFD = 90^\circ\) (tangent is perpendicular to the radius at the point of contact).
• \(\angle EGF = 90^\circ\) (angle inscribed in a semicircle, EF is a diameter), so FG ⊥ ED.

In \(\triangle EGF\) and \(\triangle EFD\): \(\angle EGF \cong \angle EFD\) (each \(90^\circ\)) and \(\angle E\) is common. So \(\triangle EGF \sim \triangle EFD\) (AA test).

EF = 2r, so \(DE \times GE = (2r)^2 = 4r^2\). Hence proved.