On the diagram above, $AB=BC=CD=DE$. The measure of $\angle ADE$ is expressed in integer degrees. Find the smallest possible $\angle ADE$.
This is what I got so far:
Let $\angle BAC=a$. So, $\angle CBD=2a$ and $\angle DCE=\angle DEC=3a$
Hence, $\angle ADE=180^\circ-a-3a=180^\circ-4a$
This was a bit iffy, but I thought that $0<\angle DCE<90^\circ$, so $0<a<30^\circ$.
From there I got $60^\circ<180^\circ-4a<180^\circ \Rightarrow 60^\circ<\angle ADE<180^\circ$. I got the smallest $\angle ADE$ as $\,61^\circ$ as a result, but it is wrong.
How do you solve this question?