Let $M = NT \cap AC$ (fig. 23). We note that since $BK$ is a bisector of $ABC$ then $\angle KBC = \angle KBA = \alpha$, and since $PT$ is parallel to $BC$ then $\angle KBC = \angle PTB$. Thus, $PT = PB = PN$. It follows that $\triangle BNT$ is right-angled with hypotenuse $BN$. Since triangle $\triangle BNK$ is also right-angled then $\angle KNT = \alpha$. Moreover, we get $\angle KAC = \angle KBC = \alpha$. Thus, quadrilateral $KANM$ is cyclic. Since $\angle ANK = 90^\circ$, then $AK$ is a diameter of the circumcircle of the $KANM$. Hence, $\angle AMK = 90^\circ$. Since $\triangle AKC$ is isosceles we get that altitude $KM$ also is a median. Hence, $AM = MC$.