Angle

ACB

is 90 degrees so we can use Pythagoras Theorem. The area of the triangle is given by

\frac{1}{2} ab = \frac{1}{2} ch

where

h

is the length of the perpendicular from

C

AB

. Then

(a + b)^{2} = a^{2} + b^{2} + 2 ab = c^{2} + 2 ch .

Thus

(a + b)^{2}

is a maximum when

h

is a maximum and equal to the radius

\frac{1}{2} c

of the circle . So the maximum value of

a + b

c \sqrt{2}

Triangle ABC inscribed in a circle with H the perpendicular from C to AB

When

AB

is not a diameter we have:

\begin{matrix} (a + b)^{2} & = a^{2} + b^{2} + 2 ab \\ = c^{2} + 2 ab \cos ∠ ACB + 2 ab \\ = c^{2} + 2 ab (1 + \cos ∠ ACB) \\ = c^{2} + 4 ab \cos^{2} \frac{1}{2} ∠ ACB . \end{matrix}

As the area of triangle ACB is given by

Δ = \frac{1}{2} ab \sin ∠ ACB

we have

\begin{matrix} (a + b)^{2} & = c^{2} + \frac{8 Δ \cos^{2} \frac{1}{2} ∠ ACB}{\sin ∠ ACB} \\ = c^{2} + 4 Δ \cot \frac{1}{2} ∠ ACB \end{matrix} .

If we keep

A

and

B

fixed and vary

C

then, as

c

and

∠ ACB

are constant, and the area of the triangle

Δ

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

AB

is a line of symmetry of the triangle. We know the radius of the circle is

c / 2 \sin ∠ ACB

(Sine Rule) and it is simple to show that, in this case,

a + b = \frac{c}{\sin \frac{1}{2} ∠ ACB} .

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 is the orthogonal bisector of the other diagonal.
3 cyclic quadrilaterals

Labeling $Q$ 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 $4 r \sqrt{2}$ and the maximum area is $2 r^{2}$ .