A Magic Triangle

As $AD || BC$ , $S_{ABC}=S_{BCD}=2$

$S$ means the area of

Using the diagram above, the area of the trapezoid $A_{ABCD} = \dfrac{1}{2}(2\sqrt{2})(\sqrt{2}a + \sqrt{2}) =$

$2a + 2$ and the area $A_{\triangle{ADC}} = \dfrac{1}{2}(\sqrt{2}a)(2\sqrt{2}) = 2a$

$\implies A_{\triangle{ABC}} = A_{ABCD} - A_{\triangle{ADC}} = 2a + 2 - 2a = \boxed{2}$ .

Let the side length of the bottom left square be $a$ and $A$ be the origin of an $xy$ -plane. Then the coordinates of the vertices of $\triangle ABC$ are $A(0,0)$ , $B(1+a, 3-a)$ , and $C(2+a, 2-a)$ . By trapezium rule the area of $\triangle ABC$ :

$\begin{aligned} [ABC] & = \left(\frac {y_B+y_A}2\right)(x_B-x_A) + \left(\frac {y_C+y_B}2\right)(x_C-x_B) + \left(\frac {y_A+y_C}2\right)(x_A-x_C) \\ & = \left(\frac {3-a+0}2\right)(1+a-0) + \left(\frac {2-a+3-a}2\right)(2+a-1-a) + \left(\frac {0+2-a}2\right)(0-2-a) \\ & = \frac {3+2a-a^2}2 + \frac {5-2a}2 - \frac {4-a^2}2 \\ & = \boxed 2 \end{aligned}$