Figure: (Show: $?= 30^{\circ}$) enter image description here 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.

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

1 Answer 1


enter image description here

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 :

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

  • $\begingroup$ Thank you sirous. $\endgroup$
    – adzetto
    Aug 16, 2022 at 9:15
  • $\begingroup$ Is it obvious that $CDEG$ is an isosceles trapezoid? $\endgroup$
    – on1921379
    Aug 16, 2022 at 9:46

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