4.4 The Midline of a Triangle
We start here with a powerful theorem about a special line in a triangle.
- Definition: A midline of a triangle is a segment whose endpoints are the midpoints of two sides of the triangle.
The Midline Theorem for triangles is useful in establishing some surprising results about quadrilaterals, as well as providing a useful theorem about trapezoids.
- Theorem 4.13:
The Midline Theorem
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half the length of the third side.
Given: $\triangle
ABC$ with midpoints $D$ and $E$ of $\overline{AB}$ and $\overline{AC}$,
respectively. Prove: $\overline{DE}\parallel\overline{BC}$ and $DE=\dfrac{1}{2}BC$ |
Proof: Given $\triangle ABD$ with midpoints $D$ and $E$. Extend $\overline{DE}$ to a point $F$ such that $DE=EF$. Since $E$ is the midpoint of $\overline{AC}$, we know that the diagonals of quadrilateral $ADCF$ bisect each other. Then Theorem 4.8 tells us that quadrilateral must be a parallelogram. The properties of a parallelogram tell us that $\overline{ADB}\parallel\overline{CF}$ and $\overline{AD}\cong \overline{CF}$. But $\overline{AD}\cong\overline{DB}$ since $D$ is the midpoint of $\overline{AB}$. Therefore $\overline{DB}\cong\overline{CF}$ and $\overline{DB}\parallel\overline{CF}$. That makes $BCFD$ a parallelogram. Now by definition, $\overline{DEF}\parallel\overline{BC}$. Also, $DF=BC$. Since $E$ is the midpoint of $\overline{DF}$, we know that $DE=\frac{1}{2}DF$, and by substitution $DE=\frac{1}{2}BC$.
In the last chapter, in our discussion of parallel lines that cut off congruent segments on one transversal, we encountered the following corollary. Since this is a partial converse of the Midline Theorem, we repeat it here.
- Corollary:
If a line goes through the midpoint of one side of a triangle and is
parallel to a second side, then the line goes through the midpoint of
the third side of the triangle.
Given: $\triangle
ABC$ with $P$ the midpoint of $\overline{AB}$ and
$\overline{PQ}\parallel\overline{BC}$. Prove: $Q$ is the midpoint of $\overline{AC}$. |
Proof: We are given $\triangle ABC$ with $P$ the midpoint of one side $\overline{AB}$ and $\overline{PQ}\parallel\overline{BC}$. Introduce line $\ell$ through $A$ such that $\ell\parallel\overline{BC}$. Now $\ell$, $\overline{PQ}$, and $\overline{BC}$ are three parallel lines that cut off congruent segments on $\overline{AB}$. The must then cut off congruent segments on any transversal--in particular, on $\overline{AC}$. This makes $Q$ the midpoint of $\overline{AC}$.
The Midline Theorem allows us to establish a variety of sometimes surprising results. One is the following fact about right triangles — the midpoint of the hypotenuse is always equidistant from all three vertices of the triangle.
- Theorem 4.14:
If $M$ is the midpoint of hypotenuse $\overline{AB}$ of right triangle
$\overline{ABC}$, then $MA=MB=MC$.
Given: Right
$\triangle ABC$ with $M$ the midpoint of hypotenuse $\overline{AB}$ Prove: $MA=MB=MC$ |
Proof: We are given that $M$ is the midpoint of hypotenuse $\overline{AB}$. Let $P$ be the midpoint of $\overline{BC}$. By Theorem 4.13, $\overline{MP}\parallel\overline{AC}$. Since $\triangle ABC$ is a right triangle with its right angle at $C$, we know that $\overline{BC}\perp\overline{AC}$. Then $\overline{BC}\perp\overline{MP}$ as well, because if a line is perpendicular to one of two parallel lines, it is perpendicular to the other. Then by definition, $\overline{MP}$ is the perpendicular bisector of $\overline{BC}$. The Perpendicular Bisector Theorem then tells us that $MB=MC$. Since $M$ is the midpoint of $\overline{AB}$, we then have $MA=MB=MC$.
This theorem appeared in the last problem set, where the approach suggested was completely different from that taken here.