Finding the area of a right angle triangle given the equations of median lines and the length of the hypotenuse

See the attached file for a rough picture. 

The slope of the line $x+y=0$ is $-1$ and the slope of the line $x+2y=0$ is $-\frac{1}{2}$. The angle $D_1$ between these two lines is given by 
\[\tan (D_1)=\frac{-1-(-\frac{1}{2})}{1+(-1)(-\frac{1}{2})}=-\frac{1}{3}  \]
\[    \Rightarrow D_1=\arctan(-\frac{1}{3})=161.565  \text{degrees}.\]
Hence 
\[D_2=180-D_1=18.435    \text{degrees}.\]
On the other hand we have
\[\tan(A_1)=\frac{1}{2}\tan (A)   \text{and}  \tan(B_1)=\frac{1}{2}\tan (B).\]
Since $A+B=90$, we have$\tan (A)\tan(B)$. From the abobe equations we have
\[\tan(A_1)\tan(A_2)=\frac{1}{4}\tan (A)\tan(B)=\frac{1}{4}\]
\[ \Rightarrow \tan(A_1)\tan(A_2)=\frac{1}{4}           (1).\]
We 
\[A_1+B_1= 90-(A_2+B_2)    \text{and}    A_2+B_2=180-D_1.\]
Thus
\[A_1+A_2=90-(180-D_1)=D_1-90=161.565-90=71.565   \text{degrees}.\]
Hence
\[3=\tan(71.565)=\tan(A_1+A_2)=\frac{\tan(A_1)+\tan(A_2)}{1-\tan(A_1)\tan(A_2)}=\frac{\tan(A_1)+\tan(A_2)}{1-\frac{1}{4}}.\]
Thus
\[\tan(A_1)+\tan(A_2)=\frac{9}{4}.   (2) \]
We can use equations (1) and (2) to solve for $\tan A_1$ and $\tan A_2$. Indeed $\tan A_1$ and $\tan A_2$ are solutions of 
\[x^2-\frac{9}{4}x+\frac{1}{4}=0   \text{or}   4x^2-9x+1=0.\]
\[\Rightarrow \tan A_1, \tan A_2=\frac{9\pm\sqrt{75}}{8}=8.830, 0.170    (3).\]
Let $x=|BC|$ and $y=|AC|$. Lets assume $\tan A_1=8.830$. Then 
\[\tan (A_1)=\frac{\frac{x}{2}}{y}=\frac{x}{2y}=8.830.\]
\[\Rightarrow x=17.66y   (4).\]
Since $x^2+y^2=36^2=1296$, we have 
\[(17.66y)^2+y^2=1296. \Rightarrow  312.8756 y^2=1296  \Rightarrow y=2.035.\]
Hence
\[ \text{x}=17.66 \times 2.035=35.94.\]
Hence the area of the triangle is
\[\frac{35.94 \times 2.035}{2}=36.56895.\]
If we assume $\tan A_1=0.170$, the above calculation will result in the exact same area for the rectangle, just the role of $x$ and $y$ will be reversed.