User:Alecscooper/Sandbox

The length of a triangle's inradius, $$r$$, is $$\frac{A}{s}$$ where $$A$$ is the area and $$s$$ is the semiperimeter.

The area of $$\triangle ABC$$ is $$\frac{AB\cdot BC}{2}$$, and it's semiperimeter is $$\frac{AB+BC+\sqrt{AB^2+BC^2}}{2}$$.

The area of $$\Box DHEI$$ is $$(AB-r)\cdot(BC-r)$$.

Combining these, $$A\Box DHEI = \left(AB-\frac{\frac{AB\cdot BC}{2}}{\frac{AB+BC+\sqrt{AB^2+BC^2}}{2}}\right)\cdot\left(BC-\frac{\frac{AB\cdot BC}{2}}{\frac{AB+BC+\sqrt{AB^2+BC^2}}{2}}\right)$$

$$=\left(AB-\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}\right)\cdot\left(BC-\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}\right)$$

$$=AB\cdot BC - \frac{(AB+BC)\cdot AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}+\left(\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}\right) ^2$$

$$A\Box ABCD - A\Box DHEI=AB\cdot r + r\cdot (BC - r)$$

$$=AB\cdot \frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}} + BC\cdot\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}-\left(\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}\right) ^2$$

$$=\frac{(AB+BC)\cdot AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}-\left(\frac{AB\cdot BC}{AB+BC+\sqrt{AB^2+BC^2}}\right) ^2$$