Find angle $x$ in the given composite figure of $\triangle BAC$ and $\triangle BDC$. A very unique question featuring a composite diagram of two triangles, with a missing angle and two equal sides.
I am posting this here to see what kind of different approaches there could be to solve it. Please feel free to leave your own answers!
(I have posted my own approach as an answer below)

 A: Yet another geometrical alternative.  Since $\angle BAC + \angle BDC = 180^\circ$, we rotate and translate $\triangle BCD$ into $\triangle FBA$ as shown in the figure below:

(Note that $A$ lies on $CF$.  However, we cannot not assume a priori that $D$ lies on $BF$, so we cannot simply observe that $\triangle BCF$ is equilateral.)
Now
$$
\angle F = \angle CBD = 180^\circ - \angle BCD - \angle BDC = 75^\circ - x.
$$
But $BC = BF$, so
$$
\angle F = \angle BCF = \angle BCD + \angle ACD = 45^\circ + x,
$$
so
$$
75^\circ - x = 45^\circ + x.
$$
Therefore, $x = 15^\circ$.
A: This is my approach to this. I'll add an explanation as well.

Here's how I go about it:
1.) Mark all the appropriate points and lines in the figure. Notice that $\angle BAC$ and $\angle BDC$ are supplementary. This gives us some motivation to construct a cyclic quadrilateral. By mapping point $B$ onto point $E$ and rotating $\triangle BDC$ about the line segment $BC$ gives us $\triangle BEC$ which is congruent to $\triangle BDC$ where $\angle BEC=105$.
2.) Notice that this not only gives us a quadrilateral $ABEC$, but in fact we get a cyclic quadrilateral as $\angle BEC + \angle BAC = 180$. Also note that segment $AB=CD=CE$. As well as $BD=BE$. Join point $A$ and $E$ via segment $AE$. Using the properties of cyclic quadrilaterals, we can see that $\angle BAE=\angle BCE=x$.
3.) Notice that quadrilateral $ABEC$ is a cyclic quadrilateral with equal non-parallel sides, this implies that $ABEC$ is in fact an isosceles trapezoid (this can also very easily be proven via some basic angle chasing and properties of congruence). This means that the diagonals of this quadrilateral are also equal, therefore segment $AE=BC$. Notice that proves that $\triangle BAE$ is congruent to $\triangle BCE$ via the SSS property. Therefore $\angle ABE=\angle BEC=105$. It follows that $\angle ACE=75$ as it is supplementary to $\angle ABE$. Therefore we can conclude that $2x+45=75$, therefore $x=15$.
A: Simple trigonometric solution.  Let $a = \lvert AB \rvert = \lvert CD \rvert$.  By the law of sines,
$$
\lvert BC \rvert = \frac{\sin 75^\circ}{\sin (x + 45^\circ)} \cdot a
$$
in $\triangle ABC$ and
$$
\lvert BC \rvert = \frac{\sin 105^\circ}{\sin (75^\circ - x)} \cdot a
$$
in $\triangle BCD$, so
$$
\sin 75^\circ \sin (75^\circ - x) = \sin 105^\circ \sin (x + 45^\circ).
$$
But $\sin 75^\circ = \sin 105^\circ$, so
$$
\sin (75^\circ - x) = \sin (x + 45^\circ).
$$
Therefore, $x = 15^\circ$.
