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.


Circle C is shown. Line segments E C and D F are radii. Lines are drawn from points E and D to point F to form chords E F and D F. Tangents E G and D G intersect at point G outside of the circle. Angle E G D is 60 degrees and angles G E C and G D C are right angles. The lengths of E F and D F are congruent.


Which statements about the arcs and angles are true? Select three options.


∠EFD ≅ ∠EGD

∠EGD ≅ ∠ECD

Arc E D is-congruent-to arc F D

mArc E F = 60°

mArc F D = 120°

Points E F and D are on circle C and angle G measures 60 The measure of arc EF equals the measure of arc FDCircle C is shown Line segments E C and D F are radii class=

Respuesta :

Answer:

∠EFD ≅ ∠EGD A

Arc ED ≅ arc FD C

m arc FD = 120° E

Step-by-step explanation:

Let us revise some facts

  1. Equal chords subtended equal arcs
  2. The measure of an inscribed angle is one-half the measure of the central angle which subtended by the same arc
  3. The measure of a central angle is equal to the measure of its subtended arc
  4. If one angle of an isosceles triangle measure 60° then the triangle is equilateral
  5. The sum of the measures of the interior angles of any quadrilateral is 360°

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 = [tex]\frac{1}{2}[/tex] m∠∠DCE

∵ m∠DCE = 120°

∴ m∠DFE = [tex]\frac{1}{2}[/tex] (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°

The relationship between the angles formed in a circle are described by

circle theorems.

The true statements are;

  • m∠EFD ≅ m∠EGD
  • [tex]\overline{ED}[/tex] ≅ [tex]\overline{FD}[/tex]
  • [tex]m\widehat{FD}[/tex] = 120°

Reasons:

According to the outside angle theorem, we have;

[tex]\displaystyle m\angle EGD = \frac{m\widehat{EFD} - m\widehat{ED}}{2}[/tex]

Therefore;

2 × m∠EGD = [tex]m\widehat{EFD} - m\widehat{ED}[/tex]

2 × 60° = 120° = [tex]m\widehat{EFD} - m\widehat{ED}[/tex]

[tex]m\widehat{EFD} + m\widehat{ED}[/tex] = 360° sum of angles formed at the center of the circle

[tex]m\widehat{ED}[/tex] = 360° - [tex]\mathbf{m\widehat{EFD}}[/tex]

120° = [tex]\mathbf{m\widehat{EFD} - m\widehat{ED}}[/tex]

120° = [tex]m\widehat{EFD}[/tex] - (360° -

120° = 2·[tex]m\widehat{EFD}[/tex] - 360°

[tex]\displaystyle \frac{120^{\circ}}{2} = \mathbf{\frac{2 \cdot m\widehat{EFD} - 360^{\circ}}{2}}[/tex]

Which gives;

60° = [tex]m\widehat{EFD}[/tex] - 180°

[tex]m\widehat{EFD}[/tex] = 60° + 180° = 240°

[tex]m\widehat{EFD}[/tex] = 240°

[tex]m\widehat{ED}[/tex] = 360° - [tex]m\widehat{EFD}[/tex] = 360° - 240° = 120°

[tex]m\widehat{ED}[/tex] = 120°

[tex]m\widehat{ED}[/tex]  = 2 × m∠EFD angle subtended by an arc at the center is twice the angle subtended at the circumference

∴ [tex]m\widehat{ED}[/tex]  = 120° = 2 × m∠EFD

[tex]\displaystyle m\angle EFD = \frac{120^{\circ}}{2} = 60^{\circ}[/tex]

Therefore;

m∠EFD = m∠EGD = m∠G = 60°

m∠EGD = 60°

  • m∠EFD ≅ m∠EGD by definition of congruency

[tex]m\widehat{ED}[/tex] = m∠ECD by definition of the measure of an arc

[tex]m\widehat{ED}[/tex] = 120° = m∠ECD

∴ m∠ECD ≠ m∠EGD = 60°

m∠ECD [tex]\ncong[/tex] m∠EGD

Given that m∠EFD = 60°, and ΔEFD is an isosceles triangle, we have;

∠DEF = ∠FDE

∠DEF + ∠FDE = 180° - 60° = 120°

∴ ∠DEF = ∠FDE = 60°

Therefore;

ΔEFD is an equilateral triangle (all angles are equal to 60°)

[tex]\widehat{ED}[/tex] = [tex]\widehat{FD}[/tex] Equal chord subtend equal minor arcs theorem

Therefore;

  • [tex]\underline{\overline{ED} \cong \overline{FD}}[/tex] by definition of congruency

[tex]\widehat{ED}[/tex] = 120° = [tex]\widehat{FD}[/tex]

  • ∴ [tex]\underline{\widehat{FD} = 120^{\circ}}[/tex]

Learn more about circle theorems here:

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