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 $\displaystyle \theta$, then the angle AOL (the B side) will be $\displaystyle 2\theta$, (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 $\displaystyle 2\theta - 180.$

Repeat the procedure for the other side of the figure and hence show that the angle KON is also $\displaystyle 2\theta-180.$

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 $\displaystyle (180-(2\theta-180))/2=180-\theta,$, in which case the angle LNK will be $\displaystyle 360-2\theta.$

Finally, the angle at C is $\displaystyle 180-\theta$ 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.

Re: Plane geometry problem

Exquisite proof!

Thank you very much :)

BTW- why don't you take a look at my second topic: http://mathhelpforum.com/number-theo...tml#post740113

Since you are so good, you may help me in solving it :P