@sandy111,
What level is this course? I can easily prove the angles are equal therefore the sides are equal, but I'd use calculus.
There are three unknown sides of the trapezoid, s1,s2 and s3 with angles a1, a2 and a3.
From law of cosines
s1^2 = r^2 + r^2 - 2r^2 cos a1 = 2r^2 (1 - cos a1)
s2^2 = 2r^2 (1 - cos a2)
Using a3 = pi - a1 - a2 and cos(pi - x) = - cos x
s3^3 = 2r^2 (1 + cos a1+a2 )
The sum of all three is 2r^2 ( 3 - cos a1 - cos a2 + cos (a1+a2))
The angle where a maximum occurs is where the partial derivative wrt a1 and a2 is zero.
The partial derivative wrt a1 = 2r^2 (sin a1 - sin(a1 + a2)) = 0
sin a1 = sin a1+a2 = 2(sin a1)(cos a2)
1 = 2 cos a2
1/2 = cos a2
a2 = pi/3
Doing the same for a2 will yield a1= pi/3 and a3 will be the same so the sides are equal and the sum is 3R. It seems like there should be a more direct way.