Angle at intersection by diagonals of a quadrilateral 
Suppose ABCD is a quadrilateral such that$ \angle BAC=50, \angle CAD=60,  \angle CBD=30$ and $\angle BDC = 25$. If $E$ is the point of intersection of $AC$ and $BD$ ,then the value of $ \angle AEB $ is __
Source

Here is a picture I've drawn of the set up,

I find that in $\triangle BCD$, we can find $ \angle BCD=125$.
I am not sure how to proceed after that. The most theorems relating angles of a quadrilateral are for cyclic quadrilaterals and this quadrilateral here is not cyclic so I am not sure how to proceed.
 A: Angles at $A$ are double of angles at $B$ or $D$:
$$\begin{align*}
\angle BAC &= 2 \angle BDC\\
\angle CAD &= 2\angle CBD
\end{align*}$$
This suggests that $A$ could be the centre of the circumcircle of $\triangle BCD$.
(The proof that $A$ is the circumcentre follows below, which involves some construction and may be less important for the multiple-choice question)
Then $AB = AD$ as radii, $\angle ADB = 35^\circ$ as a base angle. Consider the external angle of $\triangle AED$,
$$\angle AEB = \angle CAD + \angle ADB = 95^\circ$$

Extend $CA$ to point $C'_1$ where $AC'_1 = AB$. The external angle of the apex of isosceles $\triangle BAC'_1$ is $\angle BAC = 50^\circ$, so
$$\angle BC'_1C = \frac{\angle BAC}{2} = 25^\circ= \angle BDC$$
Similarly, extend $CA$ to point $C'_2$ where $AC'_2 = AD$. The external angle of the apex of isosceles $\triangle C'_2AD$ is $\angle CAD = 60^\circ$, so
$$\angle CC'_2D = \frac{\angle CAD}{2} = 30^\circ= \angle CBD$$
So both $BCDC'_1$ and $BCDC'_2$ are cyclic quadrilaterals, and both their circumcircles pass through the vertices of $\triangle BCD$, so they are all on the same circle.
$CAC'_1$ and $CAC'_2$ are on the same line, hence they are the same chord and points $C'_1 = C'_2$. So
$$AB = AC'_1 = AC'_2 = AD$$
This shows that $A$ is the circumcentre of $\triangle BDC'_1$, so is also the circumcentre of the cyclic quadrilateral $BCDC'_1$, so is also the circumcentre of $\triangle BCD$.
A: A construction procedure for the quadrilateral could be considered. You know to construct triangle BDC.
Next do you know how to construct a circular arc on a given line with given subtended angle?
If yes, then construct circular arcs  on CB,CD containing angles 50 and 60 degrees. The intersection point of these circles is A. All angles can be found at intersection, chasing one by one.
