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

Geometry: Auxiliary Lines

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

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Why do you say "buildings in the triangle" in all your geometry questions? I don't think it means whatever you think it means. – Rahul Nov 4 '12 at 5:13
@Rahul Narain solutions such as none without using trigonometry:math.stackexchange.com/questions/223893/… – felipeuni Nov 4 '12 at 5:30
That doesn't answer my question. But assuming you're Spanish, I guess when you say building you mean a construcción in geometry; this is called a geometrical construction in English. The word building is only used for a construcción in the architectural sense. – Rahul Nov 4 '12 at 5:47
Sorry for the typo. I have just edited it. – felipeuni Nov 4 '12 at 6:21

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 $f=d, x=108^\circ, d=\frac{e\sin 108^\circ\sin 6^\circ }{\sin30^{\circ}\sin36^{\circ}}=\frac{2e\sin 72^\circ\sin 6^\circ }{\sin36^{\circ}}=4e\cos 36^\circ \sin 6^\circ$

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