If $$\triangle ABC$$ is an equilateral triangle, what is the value of $$x$$?

Incorrect.
[[snippet]]
Carefully check your calculations.

Correct.
[[snippet]]
Since $$AB = AC = BC$$,
>$$\angle ABC = \angle BCA = \angle CAB = \frac{180^\circ}{3} = 60^\circ$$
Based on this,
>$$\angle ACD = 180^\circ - 60^\circ - 40^\circ = 80^\circ$$.
Given that $$\angle CAB = 60^\circ$$,
>$$\angle CAD = 90^\circ - 60^\circ = 30^\circ$$.
So,
>$$x = 180^\circ - 80^\circ - 30^\circ = 70^\circ$$.

Incorrect.
[[snippet]]
Carefully check your work.

Incorrect.
[[snippet]]
You might have gotten this answer if you assumed that $$\angle ACD$$ is a right angle.

Incorrect.
[[snippet]]
You may have gotten this answer if you tried to use the properties of parallel lines. Lines $$AD$$ and $$CE$$ are _not_ parallel.

20°

40°

60°

70°

80°