# Thread:

1. ## Plane geometry problem

Hey everyone,

Recently, I've found an interesting task from plane geometry and I have no idea how to solve it

Here it is (I translate it from my mother tongue, so sorry for any mistakes):
There is a rhomboid ABCD with acute angle at A vertex.We suppose that circumcircle on triangle ABD intersects side CB in point K and side CD in point L (K and L are different from vertexes). Segment AN is a diameter of this circle. Prove that point N is a centre of circumcircle on triangle CKL.

I add an image of this figure so that it's easier for you to understand the task:

Thanks for help in advance!
If you have any questions (e.g. my translation is unclear), feel free to post them.
Regards
Lukasz

2. ## Re: Plane geometry problem

Start by removing that line from B to D, it's distracting, but add lines from K and L to the centre of the big circle O.
Call the angle at D $\theta$, then the angle AOL (the B side) will be $2\theta$, (for angles on the same arc, the angle at the centre will be twice the angle at the circumference).
That means the angle LON will be $2\theta - 180.$
Repeat the procedure for the other side of the figure and hence show that the angle KON is also $2\theta-180.$
It follows that the triangles LON and KON are congruent in which case LN will be the same length as KN.

Since N is equidistant from L and K, it will lie on the perpendicular bisector of LK, which will be a diameter of the small circle.
The base angles of the triangles LON and KON will be $(180-(2\theta-180))/2=180-\theta,$, in which case the angle LNK will be $360-2\theta.$
Finally, the angle at C is $180-\theta$ which is double the angle LNK. Putting this with the fact that N lies on a diameter, it follows that N is the centre of the small circle.

3. ## Re: Plane geometry problem

Exquisite proof!
Thank you very much
BTW- why don't you take a look at my second topic: Prove that n is square or doubled square
Since you are so good, you may help me in solving it :P
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