Length of a side of a triangle with angle relations In the image, $AB=CD=5$, $BC=2$. Then $AD=?$
My try
I extended $BC$ until intersecting $AD$, and i extended $CD$ until intersecting $AB$. After that i tried angle chasing inside the new triangles but i got nothing. Any hints?
This problem is meant to be resolved without trigonometry.

 A: The idea of this answer is the same as the one in Lee's answer. One can get $AD$ without using trigonometry. 

Let $E$ be the intersection point of $AD$ with $BC$.
Then, since $\angle{ACE}=3x$, we get $\angle{DCE}=5x-3x=2x$.
It follows that $\triangle{CBA}\equiv\triangle{ECD}$. So, $CE=2$ and $ED=CA$.
Since $\angle{CEA}=x+2x=3x$, we see that $\triangle{ACE}$ is an isosceles triangle. It follows that $AE=AC$.
Let $AC=p$. Then, using
$$CA^2+CD^2=2(CE^2+AE^2)\tag1$$
(see here) one gets
$$p^2+5^2=2(2^2+p^2)\implies p=\sqrt{17}$$
from which
$$AD=2p=2\sqrt{17}$$
follows.

The wiki page uses trigonometry for proving $(1)$.
One can prove $(1)$ without using trigonometry.
Let $\vec{EC}=\vec c,\vec{EA}=\vec a$. Then, $\vec{ED}=-\vec a$ and
$$\begin{align}CA^2+CD^2&=|\vec{a}-\vec{c}|^2+|-\vec{a}-\vec{c}|^2
\\\\&=2|\vec a|^2+2|\vec c|^2
\\\\&=2EA^2+2EC^2\qquad\square\end{align}$$
A: This looks like a trick question:
the construction with all declared 
properties does not exist.
Let
$\angle BAC=\angle ADC=\alpha$,
$\angle CBA=\beta$, 
$E=BC\cap AD$
and 
let's ignore for the moment
that we have a condition
$\angle CBA=2\angle BAC$.
Then (as it was already noted in other answers)
we must have
\begin{align} 
\angle ECA&=\angle EAC=\alpha+\beta
,\\
|ED|&=|AC|=|AE|
,\\
|CE|&=|BC|=2
.
\end{align} 

Let $F$ be the median (and the altitude)
of the isosceles $\triangle AEC$.
Then we must 
recognize $\triangle AFB$
as the famous $3-4-5$ 
right-angled triangle,
with $|AF|=4$. 
Now we can easily find $|AE|$ and $|AD|$:
\begin{align} 
\triangle EFA:\quad
|AE|&=\sqrt{|AF|^2+|FE|^2}=\sqrt{17}
,\\
\end{align} 
and we are tricked to state that 
we have the answer:
\begin{align} 
|AD|&=2|AE|=
2\sqrt{17}
.
\end{align}
Ant it would be indeed the correct answer,
in case when 
we do not have any relations
between the angles $\alpha$ and $\beta$ .
Otherwise we must check that
the other declared conditions hold,
that is,
\begin{align} 
\angle BAC=\alpha&=x
,\\
\angle CBA=\beta&=2x
,
\end{align}
in other words
\begin{align} 
\angle CBA&=2\angle BAC
.
\end{align}
Let's check if the following is true:
\begin{align} 
\triangle AFC:\quad\alpha+\beta&
=x+2x=3x=\arctan4
,\\
\triangle AFB:\phantom{\quad\alpha+}\beta
&=2x
=
\arctan\tfrac43
.
\end{align}
So, the following must be true: 
\begin{align} 
\tfrac13\arctan4&=
\tfrac12\arctan\tfrac43
,
\end{align}
but this is false, since
\begin{align}
\tfrac13\arctan4
&\approx 0.4419392213
,\\
\tfrac12\arctan\tfrac43
&\approx 0.4636476090
,
\end{align}
thus the original question 
does not have a valid solution.
Edit
Another illustration that given geometric construction
is invalid.
Triangle $ABC$ with given constraints on the angles and side lengths
uniquely defines $x$:
\begin{align}
\triangle ABC:\quad
\frac{|BC|}{\sin x}&=
\frac{|AB|}{\sin 3x}
,\\
\frac{2}{\sin x}&=
\frac{5}{\sin x \,(3-4\sin^2x)}
,\\
\sin^2x&=\tfrac18
,\\
\sin x&=\tfrac{\sqrt2}4
,\\
x&=\arcsin\tfrac{\sqrt2}4\approx 20.7^\circ
,
\end{align}
Given $x$, we can construct $\triangle ABC$
and the point $D$, but the ray $DX:\angle XDC=x$
will miss the point $A$, 
and the real picture should in fact look like this:

A: Lets say $BC$ intersects $AD$ at point $E$, then $ABC$ is similar to $DCE$, those $AC=ED$ and $CE=2$. Also notice that $ACE$ is isosceles, thus $AC=ED=AE$, i.e. $AD=2AC$.
I don't know, but maybe it helps to solve your problem without trigonometry
