How to prove in figure that $x=30^\circ$, where we know angles $36^\circ$, $24^\circ$ and $18^\circ$?

Question

I know how to prove $\sin(156^\circ−2x)\sin24^\circ\sin18^\circ\sin36^\circ = \sin(x)\sin(x)\sin(138^\circ−x)\sin(x−12^\circ) $
but if is it possible to prove without trigonometry?

Answer 1

Geometric solution:

In figure DF is parallel with BC and we have:

$\overset{\LARGE\frown}{FDI}=24\Rightarrow \overset{\LARGE\frown}{IF}=\overset{\LARGE\frown}{ID}=48$

$\overset{\LARGE\frown}{DH}=2\times 18=36 \Rightarrow \overset{\LARGE\frown}{IH}=\overset{\LARGE\frown}{DIH}-\overset{\LARGE\frown}{ID}=36-24=12$

$\rightarrow \overset{\LARGE\frown}{FH}-\overset{\LARGE\frown}{IH}=24-12=12$

$\overset{\LARGE\frown}{CDH}=24\times 2+18\times 2=84$

$\Rightarrow \overset{\LARGE\frown}{IH}=84-12=72$

Due to statement:

$\overset{\LARGE\frown}{HB}=\overset{\LARGE\frown}{ED}\Rightarrow \overset{\LARGE\frown}{BE}=\overset{\LARGE\frown}{HD}=12+24=36$

$\overset{\LARGE\frown}{CI}=24+48=72=\overset{\LARGE\frown}{IB}\Rightarrow \overset{\LARGE\frown}{CIB}=2\times 72=144$

$\overset{\LARGE\frown}{CDE}=\overset{\LARGE\frown}{BIC}-\overset{\LARGE\frown}{BE}=144-36=108$

$\overset{\LARGE\frown}{ED}=\overset{\LARGE\frown}{CDE}-\overset{\LARGE\frown}{CD}=108-48=60$

finally:

$$x=\frac {60}2=30$$

Answer 2

Let $ABCD$ be our quadrilateral,$\measuredangle BAC=\measuredangle ADC=x$, $\measuredangle ACB=36^{\circ},$

$\measuredangle BDC=18^{\circ}$ and $\measuredangle CAD=24^{\circ}.$

Also, let $\Delta ABC\cong\Delta CB_1A$ such that $B$ and $B_1$ are placed in the same side respect to the line $AC$.

Let $B_1C\cap BD=\{K\}.$

Thus, since $\measuredangle KCA=\measuredangle KDA=x,$ we see that $AKCD$ is cyclic, which gives $$\measuredangle B_1AK=\measuredangle KAC=18^{\circ}$$ and we obtain: $$\frac{AB_1}{AC}=\frac{B_1K}{KC}$$ or $$\frac{BC}{AC}=\frac{AB-KC}{KC}$$ or $$\frac{BC}{AC}=\frac{AB}{BC}\cdot\frac{BC}{KC}-1$$ or $$\frac{\sin{x}}{\sin(144-x)}=\frac{\sin36^{\circ}}{\sin{x}}\cdot\frac{\sin24^{\circ}}{\sin(x-12^{\circ})}-1$$ or $$\frac{\cos12^{\circ}-\frac{1}{2}}{\cos12^{\circ}-\cos(2x-12^{\circ})}-1=\frac{\sin{x}}{\sin(x+36^{\circ})}$$ or $$\frac{\cos\left(2x-12^{\circ}\right)-\frac{1}{2}}{\cos12^{\circ}-\cos(2x-12^{\circ})}=\frac{\sin{x}}{\sin(x+36^{\circ})}$$ or $$\sin(3x+24^{\circ})+\sin(48^{\circ}-x)-\sin(x+36^{\circ})=$$ $$=2\sin{x}\cos12^{\circ}-\sin(3x-12^{\circ})+\sin(x-12^{\circ})$$ or $$2\sin(x-6^{\circ})\cos(2x+30^{\circ})+2\cos18^{\circ}\sin(30^{\circ}-x)=2\sin{x}\cos12^{\circ}-\sin(3x-12^{\circ})$$ or $$2\sin(30^{\circ}-x)(2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ})=$$ $$=\cos12^{\circ}(4\sin^3x-\sin{x})+\sin12^{\circ}(4\cos^3x-3\cos{x})$$ or $$2\sin(30^{\circ}-x)(2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ})=$$ $$=\cos12^{\circ}\sin{x}(4\sin^2x-1)-\sin12^{\circ}\cos{x}(4\sin^2x-1)$$ or $$\sin(30^{\circ}-x)(2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ})=\sin(x-12^{\circ})(1-2\cos2x)$$ or $$\sin(30^{\circ}-x)(2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ})+2\sin(x-12^{\circ})\sin(30^{\circ}-x)\sin(30^{\circ}+x)=0$$ or $$\sin(30^{\circ}-x)(2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ}+2\sin(x-12^{\circ})\sin(30^{\circ}+x))=0,$$ which gives $x=30^{\circ}$ because $12^{\circ}<x<78^{\circ}$ or $$2\sin(x-6^{\circ})\cos(30^{\circ}-x)+\cos18^{\circ}+2\sin(x-12^{\circ})\sin(30^{\circ}+x)=0,$$ which is $$\cos18^{\circ}+\sin24^{\circ}+\sin(2x-36^{\circ})+\cos42^{\circ}-\cos(2x+18^{\circ})=0$$ or $$\cos18^{\circ}+\sin24^{\circ}+\cos42^{\circ}=\sin(72^{\circ}-2x)-\sin(2x-36^{\circ})$$ or $$\sin(54^{\circ}-2x)=\frac{\cos18^{\circ}+\sin24^{\circ}+\cos42^{\circ}}{2\cos18^{\circ}},$$ which is impossible because: $$\frac{\cos18^{\circ}+\sin24^{\circ}+\cos42^{\circ}}{2\cos18^{\circ}}=\frac{1}{2}+\frac{\cos66^{\circ}+\cos42^{\circ}}{2\cos18^{\circ}}=$$ $$=\frac{1}{2}+\cos54^{\circ}\cdot\frac{\cos12^{\circ}}{\cos18^{\circ}}>\frac{1}{2}+\frac{1}{2}\cdot1=1.$$