Let us draw the line $l \parallel BC$ through the point $A$ and denote $W = BP \cap l$. Then

$\angle BPC = 180^\circ - (\angle PBC + \angle PCB) =$
$180^\circ - (\angle PCA + \angle PCB) = 180^\circ - \angle BCA \Rightarrow$
$\angle CPW = \angle CAW$. And so points $A, P, C, W$ are cyclic. Thus $\angle AWC = 180^\circ - \angle APC$,
$\angle APC = 180^\circ - (\angle PAC + \angle PCA) =$
$180^\circ - (\angle PAC + \angle PAB) = 180^\circ - \angle BAC$, and
it follows that (fig.17) $\angle AWC =$
$180^\circ - \angle APC = \angle BAC$. Now $AW \parallel CB$ implies
that $\angle WAC = \angle BCA$. And so $\triangle ABC \sim \triangle ACW$.
Since $M, N$ are the midpoints of the corresponding sides of similar triangles we have that
$\angle WNA = \angle AMC \Rightarrow$
$\angle WNA + \angle ANP = \angle AMC + \angle AMB = 180^\circ$. And so
$B, P, N, W$ lie on the same line.
Therefore $\angle BNA = \angle BMA$, which implies that
$A, B, M, N$ are cyclic. Since $MN \parallel AB$ as the centerline, $ABMN$ is an isosceles trapezoid, whence
$AN = BM \Rightarrow AC = BC$, what was to be proved.

Fig.17

Let $Q$ be a point of the median $AM$ such that $\angle QCB = \angle PCA$. Since $\angle QMB = \angle PNA$ we have that $\angle QMC = \angle PNC$ and so $\triangle PNC \sim \triangle QMC$. From this similarity $\frac{PC}{CQ} = \frac{CN}{CM} = \frac{BC}{AC}$ and thus $\triangle QCB \sim \triangle PCA$. Then we can obtain that (fig.18) $\angle BQC = \angle APC =$

Fig.18

$\angle PAC - \angle PCA = 180^\circ - \angle PAC + \angle PAB - \angle PCA = 180^\circ - \angle BAC$. Now let $Q'$ be symmetric to $Q$ with respect to $M$. Then $BQCQ'$ is a parallelogram and $\angle BQ'C = \angle BQC = 180^\circ - \angle BAC$. From this it also follows that the quadrilateral $ABQ'C$ is cyclic and so $\angle MAC = \angle Q'AC = \angle Q'BC = \angle OBC$. Denote $T = NP \cap BC$. Then the triangles $NTC$ and $MAC$ are similar and thus $\angle PTC = \angle NTC = \angle MAC = \angle PBC$. Points $T$ and $B$ lie on the circumcircle of the triangle $PBC$, and also on the line $BC$. Therefore $T$ coincides with one of the points $B$ or $C$. It is evident that $T$ cannot coincide with $C$, so $T=B$, which means that the points $N, P$ and $B$ lie on a line. Since $\angle ANB = \angle ANP = \angle AMB$, points $A, N, M$ and $B$ are cyclic, and it follows that $CN \cdot NA = CM \cdot CB$, which implies that $CA = CB$, i.e. that $ABC$ is an isosceles triangle.