$AB$ is a diameter of a circle and $C$ is any point on the circumference of the circle. Then, 

In $\mathrm{△}ABC$, we have

$\sin \theta =\frac{BC}{AB}$ and $\cos \theta =\frac{AC}{AB}$

$\Rightarrow BC=AB\sin \theta$ and $AC=AB\cos \theta$

Let $∆$ be the area of $\mathrm{△}ABC$. Then,

$\mathrm{\Delta }=\frac{1}{2}BC\times AC=\frac{1}{2}(AB{)}^2\sin \theta \cos \theta =\frac{1}{4}(AB{)}^2\sin 2\theta$

Clearly, it is maximum when $\sin 2\theta =1$ is maximum i.e. $\sin 2\theta =1\text{. }$In that case, $\theta =\pi /4$

$\therefore BC=AC=\frac{AB}{\sqrt{2}}$

Hence, the triangle is isoceles,