Area of triangle ABC $A = {1 \over 2} \times BC \times AM$ $= {1 \over 2} \times 2\sqrt {{r^2} - {x^2}} \times (r + x)$ $A = (r + x)\sqrt {{r^2} - {x^2}}$ ${{dA} \over {dx}} = \sqrt {{r^2} - {x^2}} - {x \over {\sqrt {{r^2} - {x^2}} }} \times (r + x)$ $= {{{r^2} - {x^2} - rx - {x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{{r^2} - rx - 2{x^2}} \over {\sqrt {{r^2} - {x^2}} }} = {{ - (x + r)(2x - r)} \over {\sqrt {{r^2} - {x^2}} }}$ ${{dA} \over {dx}} = 0 \Rightarrow x = {r \over 2}$ Sign change of ${{dA} \over {dx}}$ at $x = {r \over 2}$ $\Rightarrow$ A has maximum at $x = {r \over 2}$ $BC = 2\sqrt {{r^2} - {x^2}} = \sqrt 3 r$, $AM = r + {1 \over 2}r$ = ${3 \over 2}r$ $\Rightarrow AB = AC = \sqrt 3 r$