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
Circumcenter
- Phlembac Adib Hasan
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Re: Rectangle
The statement is not clear to me.Can you supply a figure, please ?
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Re: Rectangle
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 $\triangle ABD$.
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Re: Rectangle
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 $\triangle CKL$
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 $\triangle CKL$
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
Re: Rectangle
Ouch, sorry for my mistake. And thank you very much for your help