Is it possible that this angle is $45^\circ$?

Question

With the information given, how can $x=45^\circ$? I just could not find a way it would be possible.

Answer 1

$x$ is not $45^\circ$.

Let $E$ be the point of the intersection of $\overline{AC}$ and $\overline{BD}$.

If $x=45$, then $\angle DEC= 45^\circ$ and due to vertical angles, $\angle AEB=45^\circ$.

If $\angle AEB=45^\circ$, then $\angle ABE=45^\circ$.

But, if this were true, $\triangle ABC$ is a $45-45-90$ triangle.

If this were true, then there would be no $\triangle DEC$, as points $B$ and $E$ would lie on the circumference of the circle.

Answer 2

The question must be referring to the angle at $D$. $\hat{BDC} =45^{\circ}$.

Answer 3

$x$ is not $45°$.

But angle $BDC$ surely is.