Q. 54.2( 54 Votes )

In figure 3.81, s

Answer :

In ∆DEF,

DFE = 90° {Using tangent-radius theorem which states that a tangent at any point of a circle is perpendicular to the radius at the point of contact.}

Given: EF = diameter of the circle.

DE2 = DF2 + EF2 {Using Pythagoras theorem}

DE2 = DF2 + (2r)2

DE2 = DF2 + 4r2

DF2 = DE2- 4r2

Also, DE × DG = DF2

This property is known as tangentsecant segments theorem.

DE × DG = DE2 - 4r2

DE2- DE × DG = 4r2

DE(DE – DG) = 4r2

DE × EG = 4r2

Hence, proved.

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