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.
Best Answer
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 $\alpha=\cot^{-1} \sqrt {\frac 72}$.