## Elementary Geometry for College Students (6th Edition)

$MN=6$, $m\angle FMN=80^{\circ}$, $m\angle FNM=40^{\circ}$
$MN=6$ The mid-segment of a triangle is half the length of the base. If the base of the triangle is $12$, then the length of mid-segment $MN=6$. $m\angle FMN=80^{\circ}$ Given $m\angle FJH=80^{\circ}$. Also the mid-segment of a triangle is parallel to the base. Therefore $\angle FJH$ and $\angle FMN$ are corresponding angles. According to the Postulate 11: If two parallel lines are cut by a transversal, then the corresponding angles are congruent. Thus, $m\angle FJH=m\angle FMN=80^{\circ}$. $m\angle FNM=40^{\circ}$ The sum of the interior angles of a triangle is 180 degrees. We are given $m\angle F=60^{\circ}$ and we know that $m\angle FMN=80^{\circ}$. $360-60-80=40$ Thus $m\angle FNM=40^{\circ}$