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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