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.
 A: WARNING: This solution uses co-ordinate geometry, which had not been explicitly excluded at the time of answering.
Choose $C$ as origin, and $CB$ as $x$-axis. Then $CA$ will be the $y$-axis.
Let $A(0,a)$ and $B(b,0)$. Then, using the well-known section formula,
$$L \equiv \left(\frac {2b}{3}, \frac a3\right)$$ and $$K\equiv \left(\frac b3, \frac {2a}{3} \right)$$
Now we have:
$$CK^2=2\cdot CL^2$$
Using the distance formula, this means that:
$$\left(\frac b3\right)^2+\left( \frac {2a}{3} \right)^2=2\left( \frac {2b}{3} \right)^2+2\left( \frac a3 \right)^2$$
This simplifies to:
$$\frac ab=\sqrt {\frac 72}$$
Thus, $\frac {AC}{CB}=\tan \beta=\sqrt {\frac 72}$.
Hence, $\beta=\tan^{-1} \sqrt{\frac 72}$ and $\alpha=\cot^{-1} \sqrt {\frac 72}$.
A: Let $AK=KL=LB=x$. By the definition of cosine of an acute angle $$\cos\beta=\dfrac{a}{3x}$$ By the law of cosines in triangle $KBC$ $$\cos\beta=\dfrac{a^2+4x^2-CK^2}{4ax}$$
By the law of cosines in triangle $LBC$ $$\cos\beta=\dfrac{a^2+x^2-CL^2}{2ax}$$ So $$\dfrac{a^2+4x^2-CK^2}{4ax}=\dfrac{a^2+x^2-CL^2}{2ax}$$ Using $CK^2=2CL^2$, this can lead you to $\dfrac{a}{x}=\sqrt{2}$. This means $\cos\beta=\dfrac{\sqrt2}{3}$.
A: Let's construct median of right triangle say, $6y=CD$
$AD=BD=CD=3y$
$AK=KL=LB=2y$
$KD=DL=y$
Applying median theorem in $\triangle CKL$
$$2\times (3y)^2=x^2+2x^2-\frac{(2y)^2}{2}$$
$$x=\frac{2\sqrt{15}y}{3}$$
Applying cosine theorem in $\triangle CDL$
$\angle CDL=2\alpha$
$$\cos2\alpha=\frac{(3y)^2+y^2- (\frac {2\sqrt{15}y}{3})^2}{2 \times 3y\times y} = \frac{5}{9}$$
$$2\alpha= cos^{-1}(\frac{5}{9})=56.25$$
$$\alpha=28.13 $$
$$\beta=61.87$$
Or
$$\cos2\alpha = 2cos^2\alpha-1=\frac{5}{9}$$
$$\cos\alpha= \frac{\sqrt7}{3}$$
A: 
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  $a:b:c = \sqrt 2 : \sqrt 7 : 3$.
$m\angle A = \arcsin \dfrac{\sqrt 2}{3} \approx 28.13^\circ$.
A: Since $CL$ is a median of $\Delta KCB$, we obtain:
$$CL=\frac{1}{2}\sqrt{2CK^2+2CB^2-KB^2}$$ or
$$x=\frac{1}{2}\sqrt{4x^2+2a^2-\left(\frac{2}{3}\sqrt{a^2+b^2}\right)^2}$$ or
$$\frac{b^2}{a^2}=3.5$$ and $$\beta=\arctan\sqrt{3.5}.$$
A: 
Perform point-symmetry with respect to the point $L$. Then the image of $C$ is the point $D$ such that $D$ lies on the line $CL$ and $CL = DL$. By assumption, $BL = KL$ which means that $K$ is the symmetric image of $B$. Therefore, the quad $BCKD$ is by construction a parallelogram. However, the problem states the $$CD \cdot CL = 2\,CD \cdot CD = 2\,CD^2 = CK^2$$
which translates to
$$\frac{CL}{CK} = \frac{CK}{CD}$$ which combined with the fact that $\angle \, LCK = \angle \, KCD$ implies that the triangles $\Delta \, CKL$ and $\Delta \, CDK$ are similar. Consequently, $$\angle \, CKL = \angle \, CDK = \theta$$ But since $BCKD$ is a parallelogram, $$\angle \, LCB = \angle \, DCB = \angle \, CDK = \theta$$ Putting these angle equalities together, we get
$$\angle \, CKB = \angle\, CKL = \theta = \angle \, LCB$$
Because of the latter equality and the fact that the two triangles $\Delta \, BCK$ and $\Delta \, BLC$ share the common angle at the vertex $B$, they are similar triangles. Therefore
$$\frac{BL}{BC} = \frac{BC}{BK}$$
or alternatively
$$BK \cdot BL = BC^2$$
$$2 \, BL \cdot BL = BC^2$$
$$2 \, BL^2 = BC^2$$
Replace by $BL = \frac{AB}{3}\,$ which yields $$2 \, \left(\frac{AB}{3}\right)^2 = BC^2$$
So from there, we get
$$\sqrt{2} \frac{AB}{3} = BC$$
which becomes
$$\sin(\angle \, BAC) = \cos(\angle \, ABC) = \frac{BC}{AB} = \frac{\sqrt{2}}{3}$$
Hence
$$\angle \, BAC =  \arcsin\left( \frac{\sqrt{2}}{3} \right)$$
$$\angle \, ABC =  \arccos\left( \frac{\sqrt{2}}{3} \right)$$
