euclidean-geometryplane-geometry
(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…
$$45^{\circ}=\measuredangle EMC=\measuredangle EBC+\measuredangle ACB=\frac{1}{2}\left(\widehat{EC}+\widehat{AB}\right)=\frac{3}{2}\widehat{AB}=\frac{3}{2}\cdot\frac{360^{\circ}}{n}$$
Many thanks to @Ivan for identifying the error in the original image, and here is a solution by geometry as per the request of OP.
Extend $\small PH$ to meet $\small BC$ extend at $\small G$. $\angle CGH=53^\circ$.
$\small \therefore\angle BGP=\angle BAP\implies AGBP\ \text{is cyclic}$
As $\small AB=AP$, $\small \angle ABP=63.5^\circ$, and so is $\small \angle AGP$ because they are in the same segment.
Therefore, $\small \triangle AHG$ is a special right triangle with perpendicular sides in the ratio 1:2.
$\small \therefore HG=6$
Similarly, as $\small \triangle CGH$ is a $\small 3:4:5$ right triangle, $\small HC=8$.
Best Answer
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!