The Midpoint Theorem

The midpoint theorem is one of the most-used results in classical plane geometry. It says: in any triangle, the segment joining the midpoints of two sides is parallel to the third side and equal to half its length. Compact statement, big payoff: this theorem proves that figures are parallelograms, finds midpoints, and bisects segments , all in one step.

Statement and converse

Midpoint Theorem. In any triangle, the segment joining the midpoints of two sides is parallel to the third side and equal to half its length.

In symbols: in $\triangle ABC$, if $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$, then $DE \parallel BC$ and $DE = \tfrac{1}{2} BC$.

Converse of the Midpoint Theorem. A line drawn through the midpoint of one side of a triangle, parallel to a second side, bisects the third side.

In symbols: in $\triangle ABC$, if $D$ is the midpoint of $AB$ and $DE \parallel BC$ with $E$ on $AC$, then $E$ is the midpoint of $AC$.

Proof of the Midpoint Theorem

Given. $\triangle ABC$ with $D$ the midpoint of $AB$, $E$ the midpoint of $AC$.

To prove. $DE \parallel BC$ and $DE = \tfrac{1}{2} BC$.

Construction. Extend $DE$ to a point $F$ such that $DE = EF$. Join $FC$.

Proof.

Q.E.D.

The proof has one auxiliary construction (extending $DE$ to $F$) and uses SAS and Condition 3 of the previous lesson.

Proof of the converse

Given. In $\triangle ABC$, $D$ is the midpoint of $AB$, and $DE \parallel BC$ with $E$ on $AC$.

To prove. $E$ is the midpoint of $AC$.

Proof. Draw a line through $A$ parallel to $BC$, and extend $DE$. Since $DE \parallel BC$ and the line through $A$ is also parallel to $BC$, the line through $A$ either is the line $DE$ itself or is parallel to it. In either case, we use the alternate-interior-angle equalities and SAS or AAS in $\triangle ADE$ and a paired triangle to conclude $AE = EC$. (A cleaner argument: use similar triangles or the fact that $D$ and $E$ subdivide $AB$ and $AC$ in equal ratios, which is exactly the midpoint property.) Q.E.D.

A small army of applications

Application 1. In any quadrilateral, the midpoints of the four sides form a parallelogram.

Given quadrilateral $ABCD$ with midpoints $P, Q, R, S$ of $AB, BC, CD, DA$ respectively. By the midpoint theorem applied to $\triangle ABC$: $PQ \parallel AC$ and $PQ = \tfrac{1}{2} AC$. Applied to $\triangle ACD$: $RS \parallel AC$ and $RS = \tfrac{1}{2} AC$. So $PQ \parallel RS$ and $PQ = RS$. By Condition 3, $PQRS$ is a parallelogram.

Application 2. In a parallelogram, the diagonal bisects the other diagonal.

In parallelogram $ABCD$ with diagonals intersecting at $O$, triangle $\triangle ABD$ has $AC$ passing through the midpoint of $BD$ (which is $O$). The midpoint theorem applied to this configuration verifies bisection.

Application 3. To divide a segment $AB$ into $n$ equal parts: draw a ray from $A$, mark $n$ equal segments on it, join the last to $B$, and use the midpoint-theorem converse repeatedly.

Worked examples

Example 1. In $\triangle ABC$, $D$ and $E$ are midpoints of $AB$ and $AC$. If $BC = 10$, find $DE$.

By midpoint theorem, $DE = \tfrac{1}{2} BC = 5$.

Example 2. In $\triangle ABC$, $D$ is the midpoint of $AB$ and $DE \parallel BC$, with $E$ on $AC$. If $AC = 12$, find $AE$.

By the converse, $E$ is the midpoint of $AC$. So $AE = 6$.

Example 3. $ABCD$ is a quadrilateral with $P, Q, R, S$ the midpoints of $AB, BC, CD, DA$. Show $PQRS$ is a parallelogram.

(Application 1 above.) $PQ \parallel AC$ and $PQ = \tfrac{1}{2} AC$; $RS \parallel AC$ and $RS = \tfrac{1}{2} AC$. So $PQ \parallel RS$ and $PQ = RS$ , by Condition 3, $PQRS$ is a parallelogram.

Example 4. In $\triangle ABC$, $P$ is the midpoint of $AB$. A line through $P$ parallel to $BC$ meets $AC$ at $Q$. If $BC = 8$, find $PQ$.

By converse, $Q$ is the midpoint of $AC$. By midpoint theorem, $PQ = \tfrac{1}{2} BC = 4$.

Example 5. In $\triangle ABC$, $D, E, F$ are midpoints of $BC, CA, AB$. Find the ratio of perimeters of $\triangle DEF$ to $\triangle ABC$.

By midpoint theorem, each side of $\triangle DEF$ equals half a side of $\triangle ABC$. So perimeter of $\triangle DEF = \tfrac{1}{2} (BC + CA + AB) = \tfrac{1}{2}$ perimeter of $\triangle ABC$. Ratio: $1 : 2$.

Try it yourself

1. State the midpoint theorem.
2. State its converse.
3. Prove the midpoint theorem using SAS.
4. In $\triangle ABC$, midpoints $D, E$ of $AB, AC$. If $BC = 14$, find $DE$.
5. Show the midpoints of the sides of any quadrilateral form a parallelogram.
6. In $\triangle ABC$, $D$ is the midpoint of $AB$. A line through $D$ parallel to $BC$ meets $AC$ at $E$. Prove $E$ is the midpoint of $AC$.
7. Show that the line joining the midpoints of two sides of a triangle is parallel to the third side.
8. The midpoint theorem reduces to which special case of similar triangles?
9. Prove the converse of the midpoint theorem.
10. The diagonals of a rhombus intersect at the midpoint of each. Justify using the midpoint theorem.

Pitfalls / Insight

• Both midpoints must be involved. The theorem doesn't apply if only one endpoint is a midpoint and the other isn't.
• Parallel and half length , both pieces of information are part of the theorem.
• Converse needs parallelism + one midpoint. Without parallelism, you can't conclude.

Insight. The midpoint theorem is the cleanest bridge between midpoints and parallelism. Whenever you see midpoints in a figure, ask "does the midpoint theorem apply here?" , it usually does, and it usually shortens the proof dramatically.