Respuesta :
Answer:
- ∠EFD ≅ ∠EGD
- Arc E D is-congruent-to arc F D
- mArc F D = 120°
Step-by-step explanation:
In Quadrilateral GECD,
[tex]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[/tex]
In Triangle EFD,
[tex]\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[/tex]
Similarly, In Triangle GED
[tex]\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\°[/tex]
Finally, Arc ED is-congruent-to arc F D
The first, third and last options are true.

