Curt from Reigate College has a neat way of proving this
result by rotating one of the triangles about $C$, keeping the
other one fixed, and then using similar triangles.
It is illuminating to see different methods and you may also
like to try proving this result using the Sine Rule.
Here is Curt's method.
The question specifies the construction of a line
perpendicular to $AB$ drawn through $C$, and continued until it
intersects $DE$.
It will be proven that the perpendicular to $AB$ through $C$
bisects $DE$.
To prove this we first develop a useful tool. Note that
$\angle BCE=\pi/2$, thus if we were to rotate $B$ by $\pi/2$
radians counterclockwise to $B'$, then ECB' would be a straight
line. Also we note that $\angle ACD= \pi/2$. If we were to rotate
$A$ by $\pi/2 $ counter clockwise to $A'$, then $A'$ would coincide
with $D$ as $CD =CA=CA'$.

The diagram shows the results of rotating triangle $ABC$ by
$\pi/2$ counter clockwise.
As lengths between points are invariant under such rotations,
$CA = CA'$, $AB=A'B'$ and $BC =B'C$ thus $A'B'C$ is the
same triangle as $ABC$. Clearly $EDC$ is unaffected by the
rotation; none of its vertices were rotated or translated.
Now in order to see how this is useful, we start from the
first diagram, and draw in the extended perpendicular from $Y$ on
$AB$ passing through $C$ intersecting $DE$ at $X$. We repeat the
rotation about $C$ as shown in the next diagram.

In addition to the other transformations already discussed,
$Y$ goes to $Y'$. $XCY$ is a straight line perpendicular to $AB$,
and $Y$ is rotated $\pi/2$ about $C$ to $Y'$. Thus $XCY'$ is a
right angle and $A'B'$ is parallel to $XC$.
Again, as the distances between points that are rotated in the
same manner are invariant under rotation, it follows that
$B'Y'=BY$ and $A'Y' = AY$.
It is out intention to show that $DX = EX$, or $DE =
2DX$
