Question

Please help!

Given the following: mED=mDB=mBC


In circle F, what is the measure of EFD?


A. 17.5°

B. 35°

C. 60°

D. 70°

Answer 1

Answer: Option D.

Step-by-step explanation:

To solve this exercise you must keep on mind the Angle at the Center Theorem.

According to the  Angle at the Center Theorem, an inscribed angle is half of the central angle.

Therefore, given in the inscribed angle m∠BAC=35°, you can calculate the central angle m∠EFD as following:

$BAC=\frac{EFD}{2}$

- Solve for EFD.

$EFD=2*BAC$

- When you substitute values. you obtain:

$EFD=2(35\°)\\EFD=70\°$

Answer 2

Answer:

Option B. ∠EFD = 35°

Step-by-step explanation:

In a circle F, it has been given mED = mDB = mBC

We have to find the measure of ∠EFD

We should always remember the inscribed angle theorem which states that the measure of an inscribed angle is always half the measure of intercepted arc.

m(arc BC) = 2×∠CAB = 2×35 = 70°

Now it has been given in the question

mED = mBC

Therefore m(arc ED) = 70°

Again applying the same theorem

m(arc ED) = 2×∠EFD

70° = 2×∠EFD

m∠EFD = $\frac{70}{2}=35$

Option B. 35° is the answer.