$2\times45^\circ=90^\circ$

Hint: Fold the triangle(the creases are $AD$ and $AE$ ) so $B$ and $C$ meets at a point $F$ . You will discover that $\triangle DEF$ is a right triangle, so $DE=\sqrt{20^2+21^2}=29$

$AB=AC=x$ so $ED=\sqrt{2}x-41$ and the area sought will be $\frac{x^{2}}{2}$

By the law of cosines on each of the outer triangles

$AD^{2}=x^{2}-20\sqrt{2}x+400$ and $AE^{2}=x^2-21\sqrt{2}x+441$

Which allows the law of cosines on the inner triangle

$ED^{2}=AD^{2}+AE^{2}-2AD\cdot AE \cos{45}$

$(\sqrt{2}x-41)^{2}=x^{2}-20\sqrt{2}x+400+x^2-21\sqrt{2}x+441-2\sqrt{x^{2}-20\sqrt{2}x+400}\sqrt{x^2-21\sqrt{2}x+441}\cos{45}$ simplifies to

$x^{4}-41\sqrt{2}x^{3}+17220\sqrt{2}x-176400=0$

$(x^{2}-41\sqrt{2}x+420)(x^2-420)=0$

$x=6\sqrt{2}$ or $x=35\sqrt{2}$ or $x=2\sqrt{105}$ or $x=-2\sqrt{105}$

Since $ED > 0$ , $x>\frac{41}{\sqrt{2}}$ and the only solution that fits is $x=35\sqrt{2}$ meaning the area is $35^{2}=\boxed{1225}$