The points $A$, $M$, $P$ and $C$ lie on the same circle and $\angle MAP = \angle PAC$. Therefore, $|MP| = |PC|$ because the corresponding subtended angles are equal. Since $P$ lies on the bisector of $\overline{BC}$, we conclude that $|BP| = |CP|$. Hence $|MP| = |BP|$, which means that the point $P$ lies on the bisector of the segment $\overline{BM}$.

Let $Q$ be the midpoint of the segment $\overline{BM}$, and $O$ be the centre of the circumcircle of the triangle $ABC$.

Notice that $PQ \perp AB$ and $OM \perp AB$, which implies that $PQ \parallel OM$. Since $M$ is the midpoint of $\overline{AB}$, and $Q$ is the midpoint of $\overline{MB}$, it follows that $|MQ| = \frac{1}{3}|AQ|$. For the same reason we have $|OP| = \frac{1}{3}|AP|$.

Consider a triangle $AB_1C_1$. Its circumcentre is point $O$, while $P$ is the foot of its altitude from vertex $A$. Since $A$, $O$ and $P$ are collinear, the triangle is isosceles.

Therefore, the segment $\overline{AP}$ is also a median of the triangle, so $|OP| = \frac{1}{3}|AP|$ implies that $O$ is the centroid. Finally, since the centroid coincides with the circumcentre, the triangle $AB_1C_1$ is isosceles.