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Correct solutions from Mary of Birchwood High School and Andrei of School 205 Bucharest have contributed to the edited solution below.
As triangle DEF is equilateral, all its angles are $60$ .
Angle AEF $= 180 - 60 - c = 120 - c$
Similarly
Angle BFD $= 120 - b$
Angle EDC $= 120 - a$
From triangle FAE, I calculate angle A:
As triangle ABC is an isosceles Angle ABC $=$ Angle ACB $= x$
Therefore $180 - (120 - b + a ) = 180 - (120 - a + c )$
Therefore $60 + b - a = 60 + a - c$
Therefore $2a = b + c$
Therefore $a = (b + c)/2$
Part Two
If $a=b=c$
Angle ABC $= 180 - (120 - b + a ) = 180 - (120 - a + a ) =
60$
Angle ACB $= 180 - (120 - a + c ) = 180 - (120 - a + a ) =
60$
Therefore angle BAC is $60$
Therefore triangle ABC is equilateral.
The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?
Prove that the internal angle bisectors of a triangle will never be perpendicular to each other.