Maximise an isosceles triangle inscribed in a circle

Both copies print the complete solution and diagram. For $a>0$ and $0<\theta<\pi/2$, $AB=AC=2a\cos\theta$. Thus $A=\frac12 AB\,AC\sin2\theta=\frac12(2a\cos\theta)^2\sin2\theta=4a^2\sin\theta\cos^3\theta$. The source callout identifies $\sin2\theta=2\sin\theta\cos\theta$.

The source labels the product-rule step: $dA/d\theta=4a^2[\sin\theta\cdot3\cos^2\theta(-\sin\theta)+\cos^3\theta\cos\theta]=4a^2\cos^2\theta(\cos^2\theta-3\sin^2\theta)$.

Setting the derivative to zero gives $\cos\theta=0$ or $\tan^2\theta=1/3$. The first gives $\theta=\pi/2$, excluded by $0<\theta<\pi/2$. The second gives $\tan\theta=1/\sqrt3$ and $\theta=\pi/6$. The complete source sign table has derivative positive to the left, zero at $\pi/6$, negative to the right, with rising, horizontal and falling tangent directions. Thus the area is maximised at $\theta=\pi/6$.

At $\theta=\pi/6$, the apex angle is $2\theta=\pi/3$. Because $AB=AC$, each base angle is $(\pi-\pi/3)/2=\pi/3$. All three angles are 60°, so the triangle is equilateral. This explicit angle argument completes the requested conclusion after the source’s maximum test; it is independently authored.