Answer:
[tex]\triangle EDF[/tex] is isosceles.
Step-by-step explanation:
Please have a look at the attached figure.
We are given the following things:
[tex]\angle EDF = y[/tex]
[tex]\text{External }\angle DFG = 90 +\dfrac{y}{2}[/tex]
Let us try to find out [tex]\angle E[/tex] and [tex]\angle DFE[/tex]. After that we will compare them.
Finding [tex]\angle DFE[/tex]:
Side EG is a straight line so [tex]\angle GFE = 180[/tex]
[tex]\angle GFE[/tex] is sum of internal [tex]\angle DFE[/tex] and external [tex]\angle DFG[/tex]
[tex]\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)[/tex]
Finding [tex]\angle E[/tex]:
Property of external angle: External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.
i.e. external [tex]\angle DFG[/tex] = [tex]\angle E + \angle EDF[/tex]
[tex]\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2} -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)[/tex]
Comparing equations (1) and (2):
It can be clearly seen that:
[tex]\angle DFE = \angle E =90-\dfrac{y}{2}[/tex]
The two angles of [tex]\triangle EDF[/tex] are equal hence [tex]\triangle EDF[/tex] is isosceles.
