11.2. The median $A M$ of triangle $A B C$ divides the segment $P R$, parallel to side $A C$, with endpoints on sides $\mathrm{AB}$ and $\mathrm{BC}$, into segments of lengths 5 cm and 3 cm, starting from side $\mathrm{AB}$. What is the length of side AC?
Answer: 13 cm.

Solution. Let the ends of the segment be denoted as $\mathrm{P}$ and $\mathrm{R}$, and the point of intersection with the median $\mathrm{AM}$ as $\mathrm{Q}$, with $\mathrm{P}$ lying on side $\mathrm{AB}$ and $\mathrm{R}$ on side $\mathrm{BC}$. Draw the midline $\mathrm{MN}$ of the triangle, its length is half the length of $\mathrm{AC}$. Using the similarity of triangles $\mathrm{ANM}$ and $\mathrm{APQ}$, and $\mathrm{MAC}$ and $\mathrm{MRQ}$. The similarity ratio of the first pair is $\mathrm{k}=\mathrm{AM} / \mathrm{AQ}$, and of the second pair is $\mathrm{AM} / \mathrm{MQ}$, their sum equals one. Then $k \cdot N M=k \cdot \frac{A C}{2}=5,(1-k) \cdot A C=3$, from which $A C=13$.

Instructions. The midline is drawn and both similarities are noted: 2 points.