13.5. Given a regular triangular prism 
$A B C A_{1} B_{1} C_{1}$. Find all points on the face $A B C$ that are equidistant from the lines $A B_{1}, B C_{1}$ and $C A_{1}$.
13.5. Let $O$ be a point on the face $ABC$, equidistant from the specified lines. We can assume that $A$ is the farthest vertex from point $O$ in the base $ABC$. Consider the triangles $AOB_1$ and $BOC_1$. The sides $AB_1$ and $BC_1$ of these triangles are equal, and these are the largest sides (see problem 10.5), i.e., the bases of the heights dropped onto these sides lie on the sides themselves. Since these heights are equal, from the inequality $AO \geqslant BO$ it follows that $OB_1 \leqslant OC_1$. In the right triangles $BB_1O$ and $CC_1O$, the legs $BB_1$ and $CC_1$ are equal, so $BO \leqslant CO$.

Thus, from the inequality $AO \geqslant BO$ it follows that $BO \leqslant CO$. Continuing analogous reasoning, we get: $CO \geqslant AO$ and $AO \leqslant BO$. Therefore, $AO = BO = CO$, i.e., $O$ is the center of the equilateral triangle $ABC$.