show that of all the right triangles inscribed in a circle, the triangle with maximum perimeter is isosceles

Let ABC be the right triangle, right angled at B, inscribed inside a circle with centre O and radius r.
Let $\angle$ACB = $\theta$
Now, AC = diameter of circle = 2 r

$$\begin{gathered} In △ABC, \\ \cos \theta = \frac{BC}{AC} = \frac{BC}{2r} \\ \Rightarrow BC = 2r \cos \theta ......\left(1\right) \\ \sin \theta = \frac{AB}{AC} = \frac{AB}{2r} \\ \Rightarrow AB = 2r \sin \theta ......\left(2\right) \\ perimeter of △ABC , P = AB + BC + AC \\ \Rightarrow P = 2r \sin \theta + 2r \cos \theta + 2r \\ differentiating both sides with respect to \text{'}\theta \text{'}, we get \\ \frac{dP}{d\theta } = 2r \cos \theta - 2r \sin \theta \\ For maxima and minima, \\ \frac{dP}{d\theta } = 0 \\ \Rightarrow 2r \cos \theta - 2r \sin \theta = 0 \\ \Rightarrow \cos \theta = \sin \theta \\ \Rightarrow \frac{\sin \theta }{\cos \theta } = 1 \\ \Rightarrow \tan \theta = 1 \\ \Rightarrow \tan \theta = \tan \left({\displaystyle \frac{\mathrm{\pi }}{4}}\right) \\ \Rightarrow \theta = \frac{\mathrm{\pi }}{4} \\ \frac{d^{2}P}{d{\theta }^2 } = -2r \sin \theta - 2r \cos \theta \\ {\left[\frac{d^{2}P}{d{\theta }^2 }\right]}_{\theta = \mathrm{\pi }/4} = -2r \sin (\mathrm{\pi }/4) - 2\mathrm{r} \cos (\mathrm{\pi }/4) = -\sqrt{2}\mathrm{r} - \sqrt{2} r =-2\sqrt{2} r < 0 \\ \mathrm{So} , \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{is} \mathrm{maximum} \mathrm{at} \mathrm{\theta } = \frac{\mathrm{\pi }}{4} \\ \mathrm{now}, \mathrm{AB} = 2r\mathit{ }\sin \mathrm{\theta } = 2r \sin (\mathrm{\pi }/4) = \sqrt{2} r \\ \mathrm{BC} = 2r \cos \mathrm{\theta } = 2\mathrm{r} \cos (\mathrm{\pi }/4) = \sqrt{2} r \\ \mathrm{So}, \mathrm{AB} = \mathrm{BC} \\ \mathrm{Hence} \mathrm{perimeter} \mathrm{of} \mathrm{the} \mathrm{triangle} \mathrm{is} \mathrm{maximum} \mathrm{when} \mathrm{triangle} \mathrm{is} \mathrm{isosceles}. \end{gathered}$$