Finding Angles of a triangle

Question

In $ΔABC$, the measure of segment $AB$ is equal to the measure of segment $AC$. The $∠BAD = 30°$ . If the measure of segment $AE$ is equal to AD, what is the m∠CDE ?

What i tried: I know that $AB$ and $AC$ are the same and $AD$ and $AE$ are the same and that in $BAD$, and $A$ is 30 degrees. So I thought if angle $A$ is on the opposing side of angle $D$ , angle D would also be 30 degrees.

Answer 1

Denote $\angle ABC = \alpha$, then you can sequentially find angles $\angle ACB$, $\angle BAC$, $\angle DAE$ and finally show that $\angle AED = \alpha+15^\circ$. Since $\angle AED$ is an exterior angle, $\angle AED=\angle CDE + \angle ECD$, which allows you to find $\angle EDC$. If you did your algebra right, you should get an answer independent from $\alpha$.

Answer 2

Question is too vague. But the $\angle CDE$ will be half of $\angle BAD$, as long as both triangles $ABC$ & $ADE$ are isosceles and $\angle BAC$ is greater than $\angle BAD$. Regardless of other angles and lengths.

Check out this interactive diagram I made in geogebra for more clarity.

Answer 3

Let $|AB|=|AC|=b$, $|AD|=|AE|=d$ and $\angle ABC=\angle BCA=\beta$. Then simple angle chasing (see the picture) results in

\begin{align} x&=15^\circ . \end{align}