modifying above D F with right arrowbisects angle symbolEDG. Find FG. The diagram is not to scale.
In triangle EFG, we have ∠EGF = ∠FGD (since FG is the angle bisector of ∠EDG).
∠EGF + ∠GFE + ∠EFG = 180° (sum of angles in a triangle)
∠FGD + ∠GFE + ∠EFG = 180° (substituting ∠EGF = ∠FGD)
∠FGD + ∠FGD + ∠EFG = 180° (since ∠FGD = ∠EGF)
2∠FGD + ∠EFG = 180°
∠EFG = 180° - 2∠FGD
Since FGD and EGD are vertical angles, they are congruent.
So, ∠FGD = ∠EGD
Substituting this back into the equation:
∠EFG = 180° - 2∠EGD
Because FG is the angle bisector, we also have:
∠EFG = ∠FGD + ∠GFE
Substituting this into the equation:
∠FGD + ∠GFE = 180° - 2∠EGD
Since ∠FGD = ∠EGD, we can rewrite the equation as:
∠FGD + ∠GFE = 180° - 2∠FGD
Combining like terms:
3∠FGD + ∠GFE = 180°
The sum of the angles in triangle FGD is 180°, so:
∠FGD + ∠GFD + ∠DGF = 180°
Substituting ∠FGD = ∠EGD (since they are vertical angles):
∠EGD + ∠GFD + ∠DGF = 180°
Since FG is the angle bisector, we have ∠GFE = ∠GFD.
Substituting this back into the equation:
∠EGD + ∠GFE + ∠DGF = 180°
Since ∠EGD + ∠GFE + ∠EFG = 180°, we can rewrite the equation as:
∠EGD + ∠GFE + ∠EFG = ∠EGD + ∠GFE + ∠DGF
Canceling like terms:
∠EFG = ∠DGF
So, triangle EFG is isosceles, and EF = EG.
However, since the diagram is not to scale, we cannot determine the exact length of FG without further information or measurements.