Prove that perimeter of triangle $MNC$ is equal to half perimeter of triangle $ABC$ In $ABC$ equilateral triangle. $K$ is midpoint of $AB$. $M$ and $N$ lie on $AC$ and $BC$ respectively. If $\angle MKN=60°$, then prove that perimeter of $\triangle MNC$ is equal to half perimeter of $\triangle ABC$.
 A: 
First look at the left picture.
Mirror $N$ with respect to $CK$, let it be $N'$. We notice that $\angle CN'N=\angle MKN=60^{\circ}$. Therefore $MKNN'$ are co-cyclic. Therefore $\triangle MKN$'s mirror image with respect to $CK$ shares the same circumcircle with $\triangle MKN$. Therefore the center of $\triangle MKN$'s circumcircle lies on $CK$.
Now draw angle bisectores of $\angle CMN, \angle CNM$ and let them meet at $I$. Obviously $I$ lies on the third bisector $CK$. Since $\angle MIN=120^{\circ}$, $M,K,N,I$ are co-cyclic. Furthermore, combining with the result from previous paragraph, we know $IK$ is a diameter of that circle. Therefore $\angle IMK=\angle INK=90^{\circ}$.
Hence $MK$ bisects the outer angle $\angle AMN$ and $NK$ bisects the outer angle $\angle BNM$.
Now look at the right picture. Draw the circle tangent to $AM,MN,NB$ and let its center be $O$. We will notice that $MO$ will bisect the angle $AMN$ and $NO$ will bisect the angle $BNM$ so $O$ and $K$ are essentially the same point.
Now its easy to see the perimeter of $\triangle CMN$ is same as $CP+CQ$, which is half the perimeter of $\triangle ABC$. (Because $AP={1\over 2} AK={1\over 4}AB$ and so does $BQ$)
A: I think I have solved the problem guys!

Let's take point $P$ at side $BC$ where $\angle NKP=60°$.  Then take point $T$ at PK line where $PK=KT$. Triangles $BKP$ and $ATK$ are congruent. So $\angle TAK=60°=\angle KBP$. Notice that $AMKT$ is a circumcircle. So  $\angle TAK=\angle TMK$. Thus $TMK$ is an equilateral triangle.
Now we can be sure that triangles $MKN$ and $NKP$ are congruent. So $MN=NK$. By Ptolemy's theorem, we obtain that $AM+AT=AK$. Also, don't forget that $BP=AT$.
$CM+AM+AK=CM+2AK-AT=CM+BC-BP=CM+CP=CM+CN+NP=CM+CN+MN$.
