A geometry problem by Aly Ahmed

Let the diagonal $BD=l'$ , $BC=m'$ , and $CD=n'$ . Since $ABCD$ is a cyclic quadrilateral , $\angle CDB = \angle CAB = \theta$ , $\angle DBC = \angle DAC = \varphi$ , and $\angle BCD = 180^\circ - \angle DAB = 180^\circ -(\theta + \varphi)$ . By Ptolemy's theorem we have

$\begin{aligned} AC \cdot BD & = AB \cdot CD + BC \cdot DA \\ ll' & = mn' + nm' \\ l & = m \cdot \frac {n'}{l'} + n \cdot \frac {m'}{l'} & \small \blue{\text{By sine rule}} \\ & = m \cdot \frac {\sin \angle DBC}{\sin \angle BCD} + n \cdot \frac {\sin \angle CDB}{\sin \angle BCD} \\ & = \frac {m\sin \varphi}{\sin (180^\circ -(\theta + \varphi))} + \frac {n\sin \theta}{\sin (180^\circ -(\theta + \varphi))} & \small \blue{\text{Note that }\sin (180^\circ - \phi) = \sin \phi} \\ & = \frac {m\sin \varphi + n\sin \theta}{\sin (\theta + \varphi)} \\ \implies \frac {l\sin (\theta + \varphi)}{n\sin \theta + m\sin \varphi} & = \boxed 1 \end{aligned}$