Question

✌✌An interesting question to moderators and all here

In the attached figure, seg EF is the diameter and seg DF is a tangent segment.The radius of circle is r.

Prove that

DE × GE = 4r^2✌✌

Answer 1

${\boxed {\boxed{ \sf{Here's \: the \: answer}}}}$

$\underline { \bf{Proof :} }$

Line DF is tangent to the circle touching the circle at point F and line DGE is secant intersecting the circle at point G and E.

By Tangent - Secant segment theorem,

$\bf{DF^2 = DG \times DE}....(1)$

In ∆ DFE ,

Angle DFE = 90° ...tangent theorem

Now,
By Pythagoras theorem,

$\bf{DE^2 = DF^2 + EF^2}$

{ Diameter is twice the radius }

$\bf{ \therefore \: DE^2 = DF^2 + (2r)^2}$

$\bf{DE^2 = DF^2 + 4r {}^{2}}$

$\bf{4r {}^{2} = DE^2 - DF^2}$

From ( 1 ) ,

$\bf{4r {}^{2} = DE^2 - DG × DE}$

$\bf{4r {}^{2} = DG (DE - DG )}$

As D - G - E ,

$\bf{4r {}^{2} = DE \times GE}$

Therefore,

${\bf{DE \times GE = 4r {}^{2} }}$

Hence , Proved !

Thanks!!

Answer 2

Hey mate ^_^

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Answer:
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Given:

Seg EF is a diameter and Seg DF is a tangent segment.

Therefore,

∠HFD = $90°$

Since,

Tangent at any point of a circle is ⊥ to the radius through the point of contact.

Now,

In △ DEF,

$DE^2 = EF^2 + DF^2$ $----(1)$

By using tangent secants segments theorem, we get,

$DE × DG = DF^2----(2)$

So,

Subtract (2) from (1)

Now,

We get,

$DE^2 - DE × DG = EF^2 + DF^2 - DF^2$

$DE × (DE - DG) = EF^2$

$DE × GE = (2r)^2$

$DE × GE = 4r^2$

Hence proved,

$DE × GE = 4r^2$

#Be Brainly❤️