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