Square in a triangle! Area?

Since it is an isosceles right $\triangle$ , $AB=BC$ . $\triangle APQ \sim \triangle ABC$ , then $\dfrac{2}{AP}=\dfrac{BC}{AB}=\dfrac{AB}{AB}=1$ , so $AP=2$ . Since $\triangle ABC$ is isosceles, $SC=AP=2$ . So $AC=2+2+2=6$ . By the theorem of pythagoras, we have $AC^2=AB^2+BC^2$ , however, $AB=BC$ , so $AC^2=2AB^2$ , or $6^2=2AB^2$ . Simplifying further, we get $AB^2=18$ . Therefore, the area of $\triangle ABC$ is

$A=\dfrac{1}{2}~\text{base x height}=\dfrac{1}{2}(AB)(BC)=\dfrac{1}{2}AB^2=\dfrac{1}{2}(18)=\large{\boxed{\color{#D61F06}9}}$