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 __
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.
Best Answer
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 $\triangle BCD$.