Question

Seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is r. Prove that DE×GE=4×r×r

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)

If you still any doubts feel free to ask me.

Answer 2

Answer:

The proof is explained below.

Step-by-step explanation:

Given the circle of radius r. Seg EF is a diameter and seg DF is a tangent segment. we have to prove that $DE\times GE=4 \times r\times r$

By tangent secant theorem

$DF^{2} = DG\times DE$ →  (1)

From the given figure,

DE=GE+DG ⇒ DG=DE-GE

From 1, we get

$DF^{2}=DE\times DE - DE \times GE$

⇒$DE \times GE=DE^{2}-DF^{2}$ →  (2)

By Pythagoras theorem in ΔDEF,

$DE^{2}=DF^{2}+FE^{2}$ ⇒ $DE^{2}-DF^{2}=FE^{2}$

(2) becomes

$DE \times GE=FE^{2}$ =$(2r)^{2} = 4\times r \times r$

Hence,   $DE\times GE=4 \times r\times r$