As shown in the figure: Prove that $a^2+b^2=c^2$ Geometry: Buildings in the triangle

Other triangles with the same property:
$1.$  12  18   6  12  30 102
$2.$  15  30  15  15  15  90
$3.$  24  30  54  24   6  42
$4.$  30  10  40  30  20  50 (proposed problem in sense of clockwise)
$5.$  36  12   6  12  18  96
$6.$  36  18   6  36   6  78
$7.$  42   6  36  42  12  42
$8.$  60   6  57  30   3  24
$9.$  60  24  12  12   6  66        
Using matlab we can find all triangles (integer solutions) with this property sums of squares:


Denote the length of the lower side (the hypotenuse of the big triangle) by $h$. Then the side containing $c$ is of length $h\sin 40^\circ=h\cos 50^\circ$ (Do you see why?)
Now let's find the length of segment from the rightmost bottom vertex to the one at the bottom of $b$ (denote it by $l_1$): Using the law of sines, we get $\frac{h\sin 40^\circ}{\sin 110^\circ}=\frac{l_1}{\sin20^\circ}$, so $$l_1=\frac{h\sin 40^\circ\sin20^\circ}{\sin 110^\circ}$$
Denote by $l_2$ the length of the segment between the bottom of $b$ and the bottom of $a$. Then $\frac{l_1+l_2}{\sin50^\circ}=\frac{h\sin 40^\circ}{\sin 80^\circ}$, so $$l_2=\frac{h\sin 40^\circ\sin50^\circ}{\sin 80^\circ}-l_1=h\sin 40^\circ\left(\frac{\sin50^\circ}{\sin 80^\circ}-\frac{\sin20^\circ}{\sin 110^\circ}\right)$$
Now, we can express $a,b,c$ in terms of $h$:
$$\frac{a}{\sin30^\circ}=\frac{h-l_1-l_2}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-40^\circ))}=\frac{h-l_1-l_2}{\sin50^\circ}
\\ \frac{b}{\sin30^\circ}=\frac{h-l_1}{\sin(180^\circ-30^\circ-(180^\circ-40^\circ-70^\circ))}=\frac{h-l_1}{\sin80^\circ}
\\ \frac{c}{\sin30^\circ}=\frac{h}{\sin(180^\circ-50^\circ-30^\circ)}=\frac{h}{\sin100^\circ}$$
Now you can simplify all those expressions and substitute. (To calculate the angles, I used the fact that the sum of angles in a triangle is $180^\circ$ and each time I looked at a different triangle)
