(For reference) In the ABCD quadrilateral, $AB = 12, AD = 10, BC = 6$ and $BD =8$ Calculate $CE$, if $E$ is in $BD$ and $EB = 5$. $\angle BAD = \angle CBD$ (Answer: $4$)
My progress: I believe that the solution is by auxiliary lines... Tried to lengthen sides BC and AD forming the triangle AFB, by Geogebra it would be an isosceles triangle $\angle A = \angle F , CF =CB \implies DC \perp CB $ but I couldn't demonstrate...
In the triangle $ABD$ we have: $$\small \cos(\angle{BAD})=\frac{AB^2+AD^2-BD^2}{2\cdot AB\cdot AD}=0.75 $$
Also we know that $\small \angle{BAD}=\angle{CBD}\implies\cos(\angle{BAD})=\cos(\angle{CBD})$.
Using the very well known law of cosines, we can determine $CE$: $$\small CE^2=EB^2+CB^2-2\cdot EB\cdot CB\cdot \cos(\angle{CBD})\implies CE=4. $$
Good luck!