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:


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
$$a^2+b^2=c^2$$
