Points on the hypotenuse of a right-angled triangle Points $K$ and $L$ are chosen on the hypotenuse $AB$ of triangle $ABC$ $(\measuredangle ACB=90^\circ)$ such that $AK=KL=LB$. Find the angles of $\triangle ABC$ if $CK=\sqrt2CL$.

As you can see on the drawing, $CL=x$ and $CK=\sqrt2x$.
I don't know how to approach the problem at all. Since $\measuredangle ACB=90^\circ$, it will be enough to find the measure of only one of the acute angles. If $\measuredangle ACK=\varphi_1$ and $\measuredangle BCL=\varphi_2$, I have tried to apply the law of sines in triangle $KCL$, but it seemed useless at the end. Thank you! I would be grateful if I could see a solution without using coordinate geometry.


Draw horizontal lines from points K and L of side $\overline{AB}$ to points $K_A$ and $L_A$ of side $\overline{BC}$. Because points $K$ and $L$ trisect side $\overline{AB}$, then points $K_A$ and $L_A$ trisect side $\overline{BC}$. So $\triangle{CK_AK}$ and $\triangle{CL_AL}$ are right triangles.
Draw vertical lines from points K and L of side $\overline{AB}$ to points $K_B$ and $L_B$ of side $\overline{AC}$. Because points $K$ and $L$ trisect side $\overline{AB}$, then points $K_B$ and $L_B$ trisect side $\overline{AC}$ So $\triangle{CK_BK}$ and $\triangle{CL_BL}$ are right triangles.
We can now argue that
\begin{align}
   CL &= \sqrt 2 \; CK \\
   \sqrt{\left(\dfrac 23b \right)^2 + \left(\dfrac 13a \right)^2}
   &= \sqrt 2 \cdot \sqrt{\left(\dfrac 13b \right)^2 
                        + \left(\dfrac 23a \right)^2} \\
   a^2 + 4b^2 &= 2(4a^2+b^2) \\
   a^2 + 4b^2 &= 8a^2+2b^2 \\
   2b^2 &= 7a^2 \\
   b &= \sqrt{\dfrac{7}{2}} a
\end{align}
Since $\triangle{ACB}$ is a right triangle,
\begin{align}
   c^2 &= a^2 + b^2 \\
   c^2 &= a^2 + \dfrac 72 a^2 \\
   c^2 &= \dfrac 92 a^2 \\
   c &= \dfrac{3}{\sqrt 2} a
\end{align}
It follows that  $a:b:c = \sqrt 2 : \sqrt 7 : 3$.
$m\angle A = \arcsin \dfrac{\sqrt 2}{3} \approx 28.13^\circ$.
