Circumcenter

[Marina19]

Hello!

There is a given rectangle $ABCD$ with the acute angle next to vertex $A$. Let the circumcircle of $\triangle ABD$ cross $CB$ and $CD$ in points $K$ and $L$ (different from vertexes). Let $AN$ to be the diameter of this circle. Prove that $N$ is the center of the circumcircle of $\triangle CKL$.

Thanks in advance for help

[Phlembac Adib Hasan]

The statement is not clear to me.Can you supply a figure, please ?

[Marina19]

We have to prove that $N$ is the center of the circumcircle of $\triangle CKL$ knowing that $AN$ is the diameter of the of the circumcircle of $\triangle ABD$.

[sourav das]

I think $ABCD$ is parallelogram, not rectangle.
Proof:
$\angle ABD=\angle BDL=\angle BAL$, and since $\angle ABN = 90; \angle ADB=\angle ANB$ so, $\angle BAN = 90-\angle ADB$
It implies,
$\angle LAN=\angle LAB- \angle NAB =\angle ABD + \angle ADB -90 =90-\angle BAD$
Similarly we can show that, $\angle KAN = 90 -\angle BAD$
It implies, $arc(LN)=arc(NK)$ , so, $LN=NK$; And also, as $\angle LNK = 180-\angle LAK = 2\angle BAD=2\angle LCK$
Which implies, $N$ is the circumcenter of $\triangle CKL$

[Marina19]

Ouch, sorry for my mistake. And thank you very much for your help