Problem — Midpoints and the difference of segments

Text

On a ray with origin $P$ two points $B$ and $C$ are taken such that $PB > PC$, with $PB = 14\ \text{cm}$ and $PC = 8\ \text{cm}$. Let $M$ be the midpoint of $PB$ and $N$ the midpoint of $PC$. Prove that $MN = \dfrac{PB - PC}{2}$ and compute its measure.

Solution

Since $B$ and $C$ lie on the same ray with origin $P$ and $PB > PC$, the point $C$ lies between $P$ and $B$. The midpoints satisfy $PM = \frac{PB}{2}, \qquad PN = \frac{PC}{2},$ and, since $PN < PM$, the point $N$ also lies between $P$ and $M$. Hence $MN = PM - PN = \frac{PB}{2} - \frac{PC}{2} = \frac{PB - PC}{2}.$ This proves the required relation. $\blacksquare$

Numerical case. Substituting the values: $MN = \frac{14\ \text{cm} - 8\ \text{cm}}{2} = \frac{6\ \text{cm}}{2} = 3\ \text{cm}.$

$\boxed{MN = 3\ \text{cm}}$