Congratulations to Curt, Reigate College and to Andrei, Tudor Vianu National College, Bucharest, Romania for you solutions to this problem

When AB is a diameter angle ACB is 90 degrees so we can use Pythagoras' Theorem. The area of the triangle is given by 1/2ab = 1/2ch where h is the length of the perpendicular from C to AB. Then
(a+b)2=a2+b2+2ab=c2+2ch.
Thus (a+b)2 is a maximum when h is a maximum and equal to the radius of the circle 1/2c . So the maximum value of a+b is cÖ2.

triangle inscribed in circle with altitude h
When AB is not a diameter we have (using the Cosine Rule):
(a+b)2
= a2 +b2 + 2ab
= c2 + 2abcosÐACB +2ab
= c2 +2ab(1+cosÐACB)
=c2 +4ab cos2 1/2ÐACB.
.
As the area of triangle ACB is given by D = 1/2 absinÐACB we have
(a+b)2
= c2 + [(8Dcos2 1/2ÐACB)/(sinÐACB)]
= c2 + 4Dcot 1/2ÐACB
.
If we keep A and B fixed and vary C then, as c and Ð ACB are constant, and the area of the triangle D is a maximum when h is a maximum, it follows that a + b is a maximum when h is a maximum, that is when a=b and the altitude of the triangle drawn from C to AB is a line of symmetry of the triangle. In this case
a=b=[c
(2sin 1/2ÐACB)].

Conjecture: Let Q be a variable cyclic quadrilateral in a circle of radius r. Then the area and the perimeter of Q will be a maximum when Q is a square; that is when each of the diagonals of the quadrilateral is a diameter and each diagonal bisects of the other diagonal at right angles.

cyclic quadrilaterals
Labelling Q as ABCD, consider triangle ABC with AC fixed and B varying. Let B¢ be the position of B when, by the previous result AB + BC is a maximum, that is when AB = BC Note that this also gives the maximum area of triangle ABC. Similarly by considering triangle ADC with AC fixed, we find D¢ where AD+DC is a maximum and AD=DC. Now B¢D¢ is a diameter, keep this fixed and consider triangles B¢CD¢ and B¢AD¢. The positions of C and A that maximise the perimeter and area are C¢ and A¢ where A¢C¢ is a diameter. Hence, for the perimeter and area of Q to be a maximum all the sides of the quadrilateral must be equal making Q a square. Hence the maximum perimeter is 4rÖ2 and the maximum area is 2r2.