In concave Quadrilateral $ABCD$, prove that $AB=BC$.

Question

This is a problem I found on Facebook (the original source is mentioned within the picture). There is a concave quadrilateral ABCD such that $\angle ABD=18^\circ$, $\angle BAD=12^\circ$, $\angle CBD=54^\circ$, $\angle BCD=24^\circ$. We are asked to prove $AB=BC$.

The task is fairly simple, we need to prove that $AB=BC$, but it is trickier than it looks. On my first attempt, I tried to "complete the triangle" by joining $A$ and $C$, and then attempted to solve it by chasing some angles. However that did not lead anywhere.

I'm going to share my successful approach as an answer below. Please share your own way of proving this if you discover one!

Answer 1

Alt. hint: $\;$ by the law of sines in the two triangles

$\displaystyle\,\frac{BD}{\sin 12^\circ} = \frac{AB}{\sin 30^\circ}\,$, $\displaystyle\,\frac{BD}{\sin 24^\circ} = \frac{BC}{\sin 78^\circ}\,$, then dividing gives

$\displaystyle\, \dfrac{AB}{BC}=\dfrac{\sin 24^\circ \sin 30^\circ}{\sin 78^\circ \sin 12^\circ} = \dfrac{\sin(2\!\cdot\!12^\circ)}{2\cos12^\circ \sin 12^\circ}=1$

Answer 2

Here's my solution:

With $AB$ as a side, construct equilateral triangle $\triangle AEB$, such that $AB=EB=AB$. Notice that $\angle ADB=150^\circ$, this means that $E$ is the circumcenter of $\triangle ADB$, this implies that $AE=EB=ED=AB$. This also means that $\angle DEB=24^\circ$ and $\angle DEA=36^\circ$.

We know that $\angle DBE=78^\circ$, which means that $\angle BDE=78^\circ$ as well. Lastly, notice that $\angle BDC=102^\circ$, which means that points $E$, $D$ and $C$ are collinear. Since $\angle BEC=\angle BCE=24^\circ$, we know that $BE=BC$ which proves that $AB=BC$.