As shown in the figure: Prove that $a\,+\,b\,+\,c=d$ Geometry: Auxiliary Lines 
As shown in the figure: Prove that $a\,+\,b\,+\,c=d$

 A: 
Using sine law of triangle, $$\frac e{\sin(144^{\circ}-x)}=\frac g{\sin 36^{\circ}}$$ and $$\frac e{\sin(138^{\circ}-x)}=\frac {g+f}{\sin 42^{\circ}}$$
So, $$f=e\left(\frac{\sin 42^{\circ}}{\sin(138^{\circ}-x)}-\frac{\sin 36^{\circ}}{\sin(144^{\circ}-x)}\right)$$
$$=e\left(\frac{\sin 42^{\circ}\sin(144^{\circ}-x)-\sin 36^{\circ}\sin(138^{\circ}-x)}{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}\right)$$
$2\sin 42^{\circ}\sin(144^{\circ}-x)$
$=\cos(102^{\circ}-x)-\cos({186^\circ}-x)$ using $2\sin A\sin B$
$=\cos(102^{\circ}-x)+\cos(6^\circ-x)$ as $\cos(180^{\circ}+y)=-\cos y$
$2\sin 36^{\circ}\sin(138^{\circ}-x)=\cos(102^{\circ}-x)-\cos({174^\circ}-x)$
$=\cos(102^{\circ}-x)+\cos(6^\circ+x)$ as $\cos({174^\circ}-x)=\cos\{180^\circ-(6^\circ+x)\}=-\cos(6^\circ+x)$
So, $$f=e\frac{\cos(6^\circ-x)-\cos(6^\circ+x)}{2\sin(138^{\circ}-x)\sin(144^{\circ}-x)}=\frac{e\sin x\sin 6^\circ }{\sin(138^{\circ}-x)\sin(144^{\circ}-x)}$$
For $f=a, x=12^\circ, a=\frac{e\sin 12^\circ\sin 6^\circ }{\sin126^{\circ}\sin132^{\circ}}$
For $f=b, x=60^\circ, b=\frac{e\sin 60^\circ\sin 6^\circ }{\sin78^{\circ}\sin84^{\circ}}$
For $f=c, x=96^\circ, c=\frac{e\sin 96^\circ\sin 6^\circ }{\sin42^{\circ}\sin48^{\circ}}$
$2\sin42^{\circ}\sin48^{\circ}=\cos 6^\circ -\cos 90^\circ=\cos 6^\circ$
and $\sin 96^\circ=\sin(90+6)^\circ=\cos 6^\circ$
So,$ c=2e\sin 6^\circ $
For $f=d, x=108^\circ, d=\frac{e\sin 108^\circ\sin 6^\circ }{\sin30^{\circ}\sin36^{\circ}}=\frac{2e\sin 72^\circ\sin 6^\circ }{\sin36^{\circ}}=4e\cos 36^\circ \sin 6^\circ$
