Question

Points E, F, and D are on circle C, and angle G measures 60°. The measure of arc EF equals the measure of arc FD.

Answer 1

Answer:

• ∠EFD ≅ ∠EGD
• Arc E D is-congruent-to arc F D
• mArc F D = 120°

Step-by-step explanation:

In Quadrilateral GECD,

$60^0+90^0+90^0+ \angle ECD=360^0\angle ECD=360^0-240^0=120^0\\\angle EFD=\frac{1}{2} \angle ECD \text{ (Inscribed Angle Theorem)}\\\angle EFD =\frac{1}{2} X 120 =60^0\\Therefore: \angle EFD=\angle EGD$

In Triangle EFD,

$\angle FED =\angle FDE \text{ (base angles of an isosceles triangle)}\\\angle FED +\angle FDE+\angle EFD=180^0\\\angle FED +\angle FDE+60=180\\\angle FED +\angle FDE=120\\\angle FED =\angle FDE=60^0$

Similarly, In Triangle GED

$\angle GED =\angle GDE \text{ (tangent to a circle)}\\\angle GED +\angle GDE+\angle EGD=180^0\\\angle GED +\angle GDE+60=180\\\angle GED +\angle GDE=120\\\angle GED =\angle GDE=60^0\\Therefore, mArc F D = 120\°$

Finally, Arc ED is-congruent-to arc F D

The first, third and last options are true.

Answer 2

Answer:

1,3,5

Step-by-step explanation: