Thread: 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

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 , then the angle AOL (the B side) will be , (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
Repeat the procedure for the other side of the figure and hence show that the angle KON is also
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 , in which case the angle LNK will be
Finally, the angle at C is 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.

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