Cyclic quadrilateral

Question

A parallelogram $ABCD$ with an acute angle $BAD$ is given. The bisector of $\angle BAD$ intersects $CD$ at point $L$,and the line $BC$ at point $K$.Let $O$ be the circumcenter of $\Delta LCK$. Prove that the quadrilater $DBCO$ is inscribed in a circle.

My efforts so far:

I am halfway in solving this problem as i only need to prove that $ \angle LOD$ is equal to $y$ (see the labelings in the diagram below), since,given that the adjacent $\angle COL$ is equal to $ 2x$, I would have that $w +2x+y=180$ ,hence the quadrilateral is cyclic.

But the problem is that I really don't know how to catch angle $\angle LOD$... Any advice is appreciated.

Answer 1

Firstly, we show that $\triangle LOD\equiv\triangle COB$.

It is easy to see that $LO=CO$, $LD=DA=CB$.

Continued:

Checking the angle between the sides we have $\angle DLO=180^\circ-\angle CLO=\angle LCO+\angle LOC=\angle LCO+2x=\angle BCO$.

Since this is true, we are done as:

$\angle CDO=\angle LDO\text{ (same angle)}\\=\angle CBO\text{ (congruent triangles)}\\ \text{hence we have C, D, B, O cyclic.}$

Motivation:

You stated you wanted to find $\angle LOD$. Unfortunately, since $L$ and $D$ don't share much in common, instead of chasing that angle directly, we want to construct something so that we can move that angle. A logical way is to use $\triangle LOD$. After searching a bit, the congruence can be realised between it and $\triangle COB$.