ANSWER
Yes it is very true
<u>EXPLANATION</u>
If the two equations intersect at then this point must satisfy the two equations.
We substitute in to erquation (1)
We now substitute in to erquation (2) also
Since the point satisfy all the two equations, it is true that they intersect at
Answer:
∠EFD ≅ ∠EGD ⇒ A
Arc ED ≅ arc FD ⇒ C
m arc FD = 120° ⇒ E
Step-by-step explanation:
Let us revise some facts
In the quadrilateral CDGE
∵ m∠G = 60°
∵ m∠GDC = m∠GEC = 90°
- By using the 5th rule above
∴ m∠G + m∠GDC + m∠DCE + m∠GEC = 360°
∴ 60 + 90 + m∠DCE + 90 = 360
∴ 240 + m∠DCE = 360
- Subtract 240 from both sides
∵ m∠DCE = 120°
In circle C
∵ ∠DCE is a central angle subtended by arc DE
∵ ∠DFE is an inscribed angle subtended by arc DE
- By using the 2nd rule above
∴ m∠DFE = m∠∠DCE
∵ m∠DCE = 120°
∴ m∠DFE = (120)
∴ m∠DFE = 60°
- That means ∠EFD ≅ ∠EGD because their measure is 60°
∴ ∠EFD ≅ ∠EGD
In Δ EFD
∵ EF = FD
∵ m∠DFE = 60°
- By using the 4th rule above
∴ Δ EFD is an equilateral triangle
∴ ED = FD = FE
In circle C
∵ Side ED subtended by arc ED
∵ Side FD subtended by FD
∵ Side ED ≅ side FD ⇒ proved
- By using the 1st rule above
∴ Arc ED ≅ arc FD
∵ m∠ECD = 120°
∵ ∠ECD is a central angle subtended by arc ED
- By using the 3rd rule above
∴ m∠ECD = m arc ED
∴ m of arc ED = 120°
∵ Arc ED ≅ arc FD
∴ m arc ED = m arc FD
∴ m arc FD = 120°
