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

By tangent secant theorem,

DE×DG=DF²

DE×(DE-GE)=DF²

DE²-DE×GE=DF²

(DE²-DF²)=DE×GE ......(By Pythagoras Theorem)

EF²=DE×GE

4r²=DE×GE .......(as EF=2r)

Answer 2

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step-by-step explanation:

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,

${DE}^{2}$ = ${EF}^{2}$ + ${DF}^{2}$ ..............(i)

By using tangent secants segments theorem, we get,

DE × DG = ${DF}^{2}$ ...........(ii)

So,

Subtract (ii) from (i)

Now,

We get,

=> ${DE}^{2}$ - DE × DG = ${EF}^{2}$ + ${DF}^{2}$ - ${DF}^{2}$

=> DE × ( DE - DG ) = ${EF}^{2}$

=> DE × GE = ${2r}^{2}$

=> DE × GE = 4${r}^{2}$

Hence,

proved ✍️✍️