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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
Solution
Given
In the given figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r.
To prove, DE × GE = 4r2
seg DF ⊥ seg HF
∴ ∠HFD = 90º ...(Tangent at any point of a circle is perpendicular to the radius through the point of contact)
Using the tangent secant segments theorem, we have
DF2 = DE × DG .....(1)
Now, In Δ DFE, ∠F = 90º
By Pythagoras theorem,
DE2 = DF2 + EF2
DE2 = DE × DG + 2r2 ...(From eqn (1) and diameter = 2 × radius)
DE2 = DE × DG + 4r2
DE2 − DE × DG = 4r2
DE × (DE − DG) = 4r2
DE × GE = 4r2 ...(D - G - E)
Hence, DE × GE = 4r2
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