Exercise 92: The sum of the medians of a triangle is smaller than its perimeter but greater than its semi-perimeter.
Let there be $\triangle ABC$. Median from $\angle A$ intersects $BC$ at midpoint $D$, median from $\angle B$ intersects $AC$ at midpoint $E$, median from $\angle C$ intersects $AB$ at midpoint $F$.
$\S 48$) In a triangle, each side is smaller than the sum of the other two sides.
Median $AD$ creates $\triangle ACD$ and by triangle inequality $AC < AD + CD$.
Median $BE$ creates $\triangle ABE$ and by triangle inequality $AB < BE + AE$.
Median $CF$ create $\triangle BCF$ and by triangle inequality $BC < CF + BF$.
$AB + AC + BC < BE + AE + AD + CD + CF + BF$
$AE = \frac{AC}{2}$ ; $CD = \frac{BC}{2}$ ; $BF = \frac{AB}{2}$
$AB + AC + BC < BE + \frac{AC}{2} + AD + \frac{BC}{2} + CF + \frac{AB}{2}$
$BE + AD + CF > \frac{AB + AC + BC}{2}$
The sum of the triangle medians is greater than the semi-perimeter.