This should be easy! Right?

Refer to the image above.

We deduce that $BD=DC=6$ .

by the cosine law $20^2=12^2+13^2-2*12*13\cos(C)\Longrightarrow\cos(C)=\dfrac{-29}{104}$ Let the lenght of the sqare be $a$ Consider it to be a lenght of the triangle $6,13,a$ .It has the same angle as the big triangle! By the cosine law: $a^2=6^2+13^2-2*6*13\cos(C)=\dfrac{497}{2}\Longrightarrow a=\sqrt{\dfrac{497}{2}}$ Consider the angle opposite to the side 13 in the small triangle. let it be $A$ .Using the cosine law again: $13^2=6^2+\dfrac{497}{2}-2*6\sqrt{\dfrac{497}{2}}\cos(A)\Longrightarrow A = \arccos\left(\dfrac{11 \sqrt{\dfrac{7}{142}}}{4}\right)$ Since a square makes $\dfrac{\pi}{2}$ angle between each sides, the angle adjacent to $A$ is $\dfrac{\pi}{2}-A$ . We know the value of the two sides required for the formula of the area(the sides of a square are equal) $\text{area}=\dfrac{1}{2}*6*\sqrt{\dfrac{497}{2}}*\sin\left(\dfrac{\pi}{2}- \arccos\left(\dfrac{11 \sqrt{\dfrac{7}{142}}}{4}\right)\right)\\=\dfrac{1}{2}*6*\sqrt{\dfrac{497}{2}}*\cos\left(\arccos\left(\dfrac{11 \sqrt{\dfrac{7}{142}}}{4}\right)\right)=\dfrac{1}{2}*6*\sqrt{\dfrac{497}{2}}*\dfrac{11 \sqrt{\dfrac{7}{142}}}{4}=\dfrac{231}{8}\\ 231+8=\boxed{239}$

First we find the median $AD$ using the formula $AD=\dfrac{\sqrt{2(AC^2+AB^2)-BC^2}}{2}$ .

Then $AD=\dfrac{\sqrt{2(13^2+20^2)-12^2}}{2}=\dfrac{\sqrt{994}}{2}$ . Now let $\theta=\angle ADC$ , then $\angle CDE=90^\circ-\theta$ , also $AD=DE$ .

Using Law of cosines we get $\cos\theta=\dfrac{CD^2+AD^2-AC^2}{2\cdot CD \cdot AD}=\dfrac{6^2+\left(\dfrac{\sqrt{994}}{2}\right)^2-13^2}{2(6)\left(\dfrac{\sqrt{994}}{2}\right)}=\dfrac{77}{4\sqrt{994}}$ .

Finally, $[CDE]=\dfrac{1}{2}CD\cdot DE\cdot\sin(90^\circ-\theta)=\dfrac{1}{2}(6)\left(\dfrac{\cancel{\sqrt{994}}}{2}\right)\left(\dfrac{77}{4\cancel{\sqrt{994}}}\right)=\dfrac{231}{8}$ .

So the answer is $231+8=\boxed{239}$ .

$\color{#3D99F6}{AD^2}=\frac 1 4*(2 *AB+2*AC^2-BC^2 )= \color{#3D99F6}{\dfrac{497} 2}.\\ BD=DC=6. ~~~~~~CosADC= \dfrac{AD^2+6^2-13^2}{2*AD*6} \\ Area ~ of ~ \Delta ~CDE=\dfrac 1 2 *AD*h. ~~~But ~ h=DC*CosADC.\\ Area ~\Delta ~ CDE= \dfrac 1 2 *\cancel{AD }*DC*\dfrac{AD^2+6^2-13^2}{2*\cancel{AD*}6} .\\ Area ~\Delta ~ CDE= \dfrac 1 2 *6*\dfrac {\frac{497} 2 +6^2-13^2}{2*6} =\dfrac{231} 8. Area ~\Delta~CDE=\dfrac G H. ~~\therefore ~G + H=\Huge ~~~\color{#D61F06}{239}$