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:

 A: I add the letters to your points

Using Theorem of Sine, we get
$$\frac{a}{\sin 30^\circ}=\frac{BD}{\sin(10^\circ+40^\circ)}=\frac{BD}{\sin 50^\circ}$$
$$\frac{BD}{\sin 40^\circ}=\frac{BA}{\sin(30^\circ+20^\circ+50^\circ)}=\frac{BA}{\sin 100^\circ}$$
so we get
$$a=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\sin 40^\circ$$
Also we have
$$\frac{b}{\sin 30^\circ}=\frac{BE}{\sin(10^\circ+40^\circ+30^\circ)}=\frac{BE}{\sin 80^\circ}$$
because $\angle BAE=\angle BEA=70^\circ$, we have
$$BE=BA$$ 
so we get
$$b=\frac{BA\cdot\sin 30^\circ}{\sin 80^\circ}=\frac{BA\cdot\sin 30^\circ\cdot\sin 50^\circ}{\sin 100^\circ\cdot\sin 50^\circ}=\left(\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}\right)\cdot\cos 40^\circ$$
Finally, we have
$$\frac{c}{\sin 30^\circ}=\frac{BC}{\sin(10^\circ+40^\circ+30^\circ+20^\circ)}=\frac{BC}{\sin 100^\circ}$$
$$BC=\frac{BA}{\sin 50^\circ}$$
So, we have
$$c=\frac{BA\cdot\sin 30^\circ}{\sin 100^\circ\sin 50^\circ}$$
Since
$$\sin^2 40^\circ+\cos^2 40^\circ=1$$
So we have
$$a^2+b^2=c^2$$
A: I thought that this problem should be solvable without using trigonometry. Here's a hint for a geometric solution:

Draw the segment $DG$ and let $B'$ be the intersection of the line through $AE$ and the perpendicular line to $DG$. I claim that the triangle $DB'G$ has sides of length $a,b,c$, so $c^2 = a^2 + b^2$ by the Pythagorean theorem.
Since geometry is hard to communicate, I think it is better to let you figure this out on your own.
A: 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)
