Calculate the angle in the regular pentagon

I want calculate the angle of $$DAC$$ in the regular pentagon. But I'm not sure with my attempt. Is it right my attempt?

Since $$\angle EAB$$ divide into $$\angle EAD$$, $$\angle DAC$$, $$\angle CAB$$ and $$ABCDE$$ is regular pentagon so $$\angle EAD = \angle DAC = \angle CAB$$. We know in the regular pentagon have interior angle $$\angle EAB = 108^\circ$$, so we have $$\angle DAC=108^\circ/3= 36^\circ$$. Is it right?

• I don't understand why $\angle EAD = \angle DAC = \angle CAB$. Can you elaborate that? Commented Apr 4, 2023 at 14:44
• @Dominique I guessing it because it seems equal angle. Commented Apr 4, 2023 at 14:47
• This is not the way I see it: for me, $DEA$ is an isosceles triangle, so $\angle EAD$ and $\angle ADE$ are equal. As the third one is $108°$, the others should be $\frac{180-108}{2}$, so you get $36°$. From there you can derive that $\angle DAC$ is also $36°$, so the three angles being equal is a consequence, not a cause. (At least that's how I see it) Commented Apr 4, 2023 at 14:53

Your solution is correct. Perhaps an even easier way to see this is to note that central angle $$\angle DOC$$ measures $$\frac{360^\circ}{5} = 72^\circ$$ due to symmetry, and then $$\angle DAC$$ is the corresponding inscribed angle, so it measures $$\frac{1}{2} (72^\circ) = 36^\circ$$.
Recall too that constructing a regular pentagon requires first constructing an isosceles triangle in which each base angle is double the angle at the vertex (Euclid, Elements, IV, 10). This is $$\triangle DAC$$, so $$\angle DAC=36^o$$.