Question

If A is the center of the circle, then which statement explains how segment ED is related to segment FD?

Answer 1

Answer:

ED and FD are congruent segment

Step-by-step explanation:

we know that

Triangles A.D.E and A.D.F are right triangles ----> given problem

In the right triangle A.D.E

Applying the Pythagoras theorem

$A.E^{2}=ED^{2}+D.A^{2}$

we have

$E.D=r\ units$ -----> the radius of the circle

substitute and solve for E.D

$r^{2}=E.D^{2}+D.A^{2}$

$E.D^{2}=r^{2}-D.A^{2}$ -----> equation A

In the right triangle A.D.F

Applying the Pythagoras theorem

$A.F^{2}=F.D^{2}+D.A^{2}$

we have

$A.F=r\ units$ -----> the radius of the circle

substitute and solve for F.D

$r^{2}=F.D^{2}+D.A^{2}$

$F.D^{2}=r^{2}-D.A^{2}$ -----> equation B

Compare equation A with equation B

$E.D^{2}=r^{2}-D.A^{2}$

$F.D^{2}=r^{2}-D.A^{2}$

therefore

ED≅FD

ED and FD are congruent segment