National Olympiad Final Round

Let $N$ be the intersection point of the bisector of angle $BAC$ and side $BC$; it suffices to prove that $KN=MN$ (Fig. 19). The bisector property implies $\frac{NC}{NB}=\frac{AC}{AB}$. Substituting $NC=BC-NB$ gives $$NB=\frac{BC}{1+\frac{AC}{AB}}=\frac{AB\cdot BC}{AB+AC}.$$ As $AB+AC=2BC$ by assumption, this implies $NB=\frac{AB}{2}$.

Fig. 19 Fig. 20

On the other hand, let $X$ and $Y$ be the points of tangency of the incircle of triangle $ABC$ with sides $AB$ and $AC$, respectively (Fig. 20). Then $AX=AY$, $BX=BK$ and $CK=CY$, whence $BC=BK+CK=BX+CY=AB-AX+AC-AY=AB+AC-2AX=2BC-2AX$. Thus $BC=2AX$, implying $AX=BM$. Consequently, $KN=BN-BK=BN-BX=\frac{AB}{2}-(AB-AX)=AX-\frac{AB}{2}=BM-\frac{AB}{2}=BM-BN=MN$.