Find the length measure $x$ in right triangle $\triangle ABC$ As title suggests, the objective is to solve for the missing length $x$ in this problem. I spent some time on the problem and figured out, what I believe, a very simple approach, I'll post it as an answer down below, please share your own approaches as well!

 A: Let $E$ be the point on $BC$ directly below $A$, and let $y=BE$.  So $2y=AE=ED$.
$EC=2AE \implies y=2$.
$\text{Area}_{\triangle{ABC}}=\frac{1}{2}(AB)(AC)=\frac{1}{2}(AE)(BC) \implies x=2\sqrt{5}$.
A: This is my own approach. I'll add the explanation below as well!

Here's how I go about it:
1.) Locate a point $E$ on segment $AC$ and connect it to points $D$ and $B$ via $DE$ and $BE$ respectively, such that $\angle BEA=45$. Notice that the inscribed angles from the segment $AB$ are the same ($45$), therefore we can conclude that the quadrilateral $AEDB$ is in fact cyclic. Note that, in $\triangle BAE$, the $\angle BAE=90$ and $\angle BEA=45$, therefore $\angle EBA=45$, as well, this proves that $\triangle BAE$ is an isosceles right triangle.
2.) Above implies that $\angle ADE=45$ via the properties of cyclic quadrilaterals, which means that the line segment $DE$ is perpendicular to segment $BC$. This gives us another right triangle $\triangle DEC$ that is in fact similar to the $\triangle ABC$ via the AAA property. We can label the segment $DE$ as a temporary variable $a$, computing this will help us compute $x$.
3.) Since $\triangle DEC$ is similar to $\triangle ABC$, we can apply the Basic Proportionality Theorem and solve for both $a$, and eventually $x$. Therefore, we can write:
$$\frac{a}{x}=\frac{4}{2x}$$
Therefore, $a=2$.
We can now apply the Pythagorean theorem in $\triangle DEC$ to get $x$, therefore $x=2\sqrt {5}$
A: As $\sin C=\frac{1}{\sqrt{5}}$ and $\cos C=\frac{2}{\sqrt{5}}$ we have $\sin \angle DAC=\sin(45-C)=\sin45\cos C-\cos45\sin C=\frac{1}{\sqrt{10}}.$ Then by law of sines on the triangle $\triangle DAC$, $\frac{2x}{\sin 135}=\frac{4}{\sin \angle  DAC}$, we get $2\sqrt{2}x=4\sqrt{10}$ and $x=2\sqrt{5}$.
