A problem with isosceles triangle and unknown angles

Question

I'm struggling with this problem:

Let there be a triangle $ABC$, $AB = AC$. Angle that corresponds to corner $A$ is bigger than $30^\circ$. Let point $D$ be the point on the side $BC$ such that $\angle{BAD} = 30^\circ$ and let point $E$ be the point on the side $AC$ such that $AE = AD$.

I need to find the measure of $\angle EDC$.

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What is the fastest solution to this problem?

Answer 1

Let $\angle DAC = x$. Since $AE=AD$, we must have $\angle AED = \angle ADE = 90 - \frac{x}{2}$. So $\angle CED = 180 - \angle AED = 90 + \frac{x}{2}$. Furthermore, since $AB = AC$ with $\angle BAC = 30 + x$, we have $\angle ACB = 90 - \frac{30+x}{2} = 75 - \frac{x}{2}$. We finally compute $\angle EDC = 180 - \angle CED - \angle ACB = 180 - (90 - \frac{x}{2}) - (75 - \frac{x}{2}) = 15$.

Answer 2

Let $\angle ACB=x$ and $\angle AED=y$.

Then $\angle ABC=x$ and $\angle BAC=180^\circ-2x$.

$\angle ADE=y$ and $\angle DAE=180^\circ-2y$.

$\angle BAC-\angle DAE=\angle BAD=30^\circ$ $\implies$ $(180^\circ-2x)-(180^\circ-2y)=30^\circ$ $\implies $y-x=15^\circ$.

$\angle EDC=\angle AED-\angle ACB=y-x=15^\circ$.