A while back, this question was asked on MSE:
In fact, allow me to phrase the problem in a slightly different manner:
In quadrilateral $ABCD$, $AB=6$, $\angle{ABC}=90°$, $\angle{BCD}=45°$ and $\angle{CAD}=2\angle{ACB}$. If $DE$ is perpendicular to $AC$ with $E$ on side $BC$, prove that the length of $CE=12$.
I have managed to prove the above result, but was unable to avoid the use of some trigonometry and algebraic manipulations.
My solution is as follows:
Let $M$ be the point of intersection of line segments $AC$ and $DE$, and let $H$ be the foot of the perpendicular from $M$ to line segment $EC$. Also, let $BC=x$, $CE=a$. Finally, let $\angle ACB =\theta, \angle CAD = 2\theta, \angle ACD=45^{\circ}-\theta$.
By Pythagoras' Theorem, $AC=\sqrt{AB^2+BC^2}=\sqrt{36+x^2}$. Clearly, $\triangle{CME} \sim \triangle{CBA} \Rightarrow \frac{CM}{CE}=\frac{BC}{AC} \Rightarrow CM=CE \cdot \frac{BC}{AC}=\frac{ax}{\sqrt{36+x^2}}$.
Thus $AM=AC-MC=\sqrt{36+x^2}- \frac{ax}{\sqrt{36+x^2}}=\frac{36+x^2-ax}{\sqrt{36+x^2}} \Rightarrow \frac{CM}{AM} = \frac{ax}{36+x^2-ax}$. Now, $\tan(2\theta)=\frac{MD}{MA}, \tan(45^{\circ}-\theta)=\frac{MD}{MC} \Rightarrow \frac{\tan(2\theta)}{\tan(45^{\circ}-\theta)}=\frac{MC}{MA}=\frac{ax}{36+x^2-ax}$.
On the other hand, $\tan(\theta)=\frac{AB}{BC}=\frac{6}{x} \Rightarrow \tan(2\theta)=\frac{2\tan(\theta)}{1-\tan^2(\theta)}=\frac{2 \cdot \frac{6}{x}}{1-\frac{36}{x^2}}=\frac{12x}{x^2-36} $. Also, $\tan(45^{\circ}-\theta)=\frac{\tan(45^{\circ})-\tan(\theta)}{1+\tan(45^{\circ})\tan(\theta)}=\frac{1-\tan(\theta)}{1+\tan(\theta)}=\frac{1-\frac{6}{x}}{1+\frac{6}{x}}=\frac{x-6}{x+6} \Rightarrow \frac{\tan(2\theta)}{\tan(45^{\circ}-\theta)} = \frac{12x}{(x-6)^2}$.
Thus, we have $\frac{12x}{(x-6)^2}=\frac{ax}{36+x^2-ax} \Rightarrow a= (36+x^2-ax) \cdot \frac{12}{(x-6)^2} \Rightarrow a[1+\frac{12x}{(x-6)^2}]= 12 \cdot \frac{36+x^2}{(x-6)^2} \Rightarrow a \cdot \frac{x^2+36}{(x-6)^2} = 12 \cdot \frac{36+x^2}{(x-6)^2} \Rightarrow a=12$.
But this solution is, admittedly, rather tedious. Thus, I wonder if there exists a synthetic solution by any chance?
Let $F$ on $CD$ so that $AF\parallel BC$ and $H$ on $AC$ so that $FH\perp CD$.
Since $AF\parallel BC$, it follows that
$$\angle CAF = \angle ACB = \frac{\angle CAD}{2}.$$
Thus $\angle CAF = \angle DAF$. Also since $AF\parallel BC$,
$$\angle AFD = \angle BCD = 45^\circ.$$
Therefore $\angle AFD = \angle AFH$. It follows that $D$ and $H$ are symmetry along $AF$, and that $DH\perp CE$. Since $CH\perp DE$, $H$ is the orthocenter of $\triangle CDE$, and so $E, F, H$ are colinear.
Finally, $\triangle EFC$ has $\angle EFC=90^\circ, \angle ECF = 45^\circ$, so $CE$ is twice the distance from $F$ to $CE$ and is $12$.