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

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

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.