$\triangle ABC$ and a circle $k(O; d=AB)$

Question

$\triangle ABC$ is an equilateral triangle. Circle $k(O; d=AB)$ is given. $k$ $\cap$ $CO=\{P;Q\}$ such that $P$ is between $C$ and $O$. $PK\perp MC$ $(K \in CM)$ and $QM\perp CM$ $(M \in CM)$. $k$ intersects $CA$ in $L$ $(L \in$ $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AP})$. Show that:

1. $KL=MA, CL=BO$ and $KM=BL$
2. If $k$ $\cap$ $BC=R$, $LORC$ is a circumscribed quadrilateral.

I am having problems with the first part of the problem. I will show you what I have done. $\angle ALB = \dfrac{1}{2}\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}\arc{AB}=\dfrac{1}{2}180^\circ=90^\circ$ (inscribed angle). $BL$ is an altitude but triangle $ABC$ is equilateral, thus $BL$ is also a mediаn. $CL=\dfrac{1}{2}AC$ and $BO = \dfrac{1}{2}AB$, thus $CL=BO$.

$MA=ML-AL$ and $LK=AK-AL$. If we show $ML=LK$, we can say $MA=LK$, but I don't see it. Thank you in advance!

Answer 1

Hint: Set $AO = x$, and by repetitively applying the ratio of the sides of the right-angle triangle with angle $30˚$ and $60˚$ being $1:\sqrt{3}:2$, you can obtain the expression of all the segments in the graph with $x$. Then by comparing the ratio of the coefficient of $x$, part 1 can be solved.

Answer 2

Observe that $\angle KAP$ and $\angle MAQ$ are complementary angles. Conclude, by ASA criterion, that $\triangle APK\cong \triangle AMQ$.

Thus $MQ \cong AK$.

Now use the fact that $\angle QLB = 45^\circ$ (because it subtends chord $QB$), to show that $\triangle MLQ$ is isosceles.

The thesis follows, by following the path you suggest.