# Find a synthetic solution: $T(12^{\circ}, 24^{\circ}, 54^{\circ}, 18^{\circ}, 30^{\circ})$

Figure: (Show: $$?= 30^{\circ}$$) I was able to solve this using the trigonometric ceva. $$\frac{\sin{(72^{\circ}-\alpha)}}{\sin{(12^{\circ})}}\frac{\sin{(24^{\circ})}}{\sin{(54^{\circ})}}\frac{\sin{(18^{\circ})}}{\sin{(\alpha)}}=1$$ But I couldn't find a synthetic solution. For this, I tried to choose a point $$E$$ with $$AB=AE$$ on the $$[BC$$ line. $$AC=CE$$ and $$\angle{AFC}=\angle{CFE}=30^{\circ}$$ can be found if we construct an equilateral triangle $$AEF$$ below $$[AE]$$. But after that I couldn't get anything. Any help will be appreciated.

• Note that $\angle C = \angle A = 72^\circ$, so this is an isosceles triangle. Maybe you can use the symmetry somehow Aug 16, 2022 at 4:15
• @gt6989b I noticed, but unfortunately I couldn't get anything. Aug 16, 2022 at 4:19

Hint: Triangle ABC is isosceles and $$\angle BAC=72^o$$. As can be seen in figure the bisector of angle ACB meets side AB at E . We have $$\angle ECD=\angle ACD=18^o$$. Triangle ACE is isosceles(why ?), therefore CD is altitude and triangle ADE is isosceles. We draw the circumcircle of triangle CDE, it intersect DB at G. . We have:

$$\angle EGD=\angle ECD=18^o$$

Quadrilateral CDEG is isosceles trapezoid(why?), so triangle DEF is isosceles. By angle chasing we find $$\angle FED=30^o$$ which results in:

$$\angle DEA=42^o=\angle EAD$$

Which finally gives :

$$\angle DAC=72-42=30^o$$

• Thank you sirous. Aug 16, 2022 at 9:15
• Is it obvious that $CDEG$ is an isosceles trapezoid? Aug 16, 2022 at 9:46