Solution:

If the quadrilateral has an inscribed circle then $AM + PN = AN + PM$ (consider the tangents to the circle from $A, M, P, N$). But if $AB > AC$, then $BM > CN$ (see below). We have $AN = AB / 2$, $PM = BM / 3$, $AM = AC / 2$, $PN = CN / 3$, so it follows that $AM + PN < AN + PM$. Similarly, $AB < AC$ implies $AM + PN > AN + PM$, so the triangle must be isosceles.

To prove the result about the medians, note that $BM^2 = BC^2 + CM^2 - 2 BC \cdot CM \cos C = (BC - CM \cos C)^2 + (CM \sin C)^2$. Similarly, $CN^2 = (BC - BN \cos B)^2 + (BN \sin B)^2$. But $MN$ is parallel to $BC$, so $CM \sin C = BN \sin B$. But $AB > AC$, so $BN > CM$ and $B < C$, so $\cos B > \cos C$, hence $BN \cos B > CM \cos C$ and $BC - CM \cos C > BC - BN \cos B$. So $BM > CN$.