Book A Free Math Class

# Midpoint Theorem - Statement, Proof, and Examples

## TL;DR

The midpoint theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and exactly half its length. This article gives the formal statement, a full labelled proof, the converse, six worked examples, and the common mistakes, plus how the theorem relates to the triangle's midsegment.

## A Shortcut That Measures A Distance You Can't Reach

Suppose you need the width of a river but can only walk one bank. Mark the midpoints of two survey lines running back from the water, join them, measure that short segment, and double it - you now have the width without crossing. That trick is the midpoint theorem in action: join two midpoints, and the segment you draw is always exactly half the far side, and always parallel to it.

## What Does The Midpoint Theorem State?

The **midpoint theorem** states: in any triangle, the line segment joining the **midpoints of two sides** is **parallel to the third side** and **equal to half its length**. Formally, in △ABC, if D is the midpoint of AB and E is the midpoint of AC, then DE∥BC and DE=1/2 BC.

A _midpoint_ is the point that divides a segment into two equal halves; the [midpoint formula](/content/math/geometry/midpoint-formula/index.html) locates it from coordinates. The theorem delivers two conclusions at once, and both matter: a **direction** result (the new segment runs parallel to the third side) and a **length** result (it is exactly half as long). Missing either half means missing the theorem.

The segment DE itself has a name: it is a **midsegment** of the triangle. This page focuses on the theorem - its statement and proof - while the companion page on the [midsegment of a triangle](/content/math/geometry/midsegment-of-a-triangle/index.html) focuses on the segment and its properties. They describe the same figure from two angles: the theorem is the rule, the midsegment is the object the rule is about.

## How Is The Midpoint Theorem Proved?

The proof is worth doing once, because seeing _why_ DE is half of BC makes the result stick far better than memorising it. The strategy: extend DE to build a parallelogram, then let the parallelogram's own properties finish the job. The labelled diagram below tracks every point the proof names.

**Given:** In △ABC, D and E are the midpoints of AB and AC, so AD=DB and AE=EC.

**To prove:** DE∥BC and DE=1/2 BC.

**Construction:** Extend DE to a point F such that EF=DE. Join C.

**Proof, step by step:**

In △AED and △CEF:

AE=EC 
∠AED=∠CEF (vertically opposite angles)
DE=EF (by construction)

So △AED≅△CEF by the SAS congruence rule.

Therefore AD=CF and ∠ADE=∠CFE (corresponding parts of congruent triangles).

Because ∠ADE=∠CFE are equal alternate angles, AD∥CF, which means DB∥CF.

Now AD=DB (D is a midpoint) and AD=CF (just shown), so DB=CF.

Since DB∥CF and DB=CF, the quadrilateral BCFDB is a parallelogram.

In a parallelogram opposite sides are equal and parallel, so:

DF∥BC, hence DE∥BC.

DF=BC.

But DF=DE+EF=2DE, so 2DE=BC, giving DE=1/2 BC.

Both conclusions are proved: DE∥BC and DE=1/2 BC.

## What Is The Converse Of The Midpoint Theorem?

The **converse** runs the theorem backward: the line drawn through the midpoint of one side of a triangle, parallel to a second side, **bisects the third side**. In △ABC, if D is the midpoint of AB and a line through D is parallel to BC, that line meets AC at its midpoint E.

A compact way to hold both directions:
- **Midpoint theorem:** midpoint + midpoint → parallel and half.
- **Converse:** midpoint + parallel → the other midpoint.

The converse is what lets you _prove_ a point is a midpoint using only a parallel line - no measuring required. It is the tool behind many constructions and coordinate-geometry proofs involving [similar triangles](/content/math/geometry/similar-triangles/index.html).

## Examples of the Midpoint Theorem

Six examples, from a one-step length to a full quadrilateral argument.

### Example 1

**In △ABC, D and E are midpoints of AB and AC. If BC = 10 cm, find DE.**

By the midpoint theorem, DE=1/2 BC.

DE=1/2 × 10

DE=5 cm

Final answer: 5 cm.

### Example 2

**In △PQR, M and N are midpoints of PQ and PR. A student measures MN = 6 cm and concludes QR = 6 cm. What went wrong?**

The intuitive move is to treat the midsegment and the third side as equal since both look like "the bottom" of a smaller and larger triangle sitting together.

That skips the halving. The theorem says the midsegment is _half_ the third side, not equal to it. Reading MN=QR drops the factor of 1/2.

The correct method doubles the midsegment to recover the third side:

MN=1/2 QR

QR=2×MN=12 cm

Final answer: 12 cm. The midsegment is the _half_, so the third side is the _double_ - never equal.

### Example 3

**In △ABC, DE joins midpoints of AB and AC with DE = 4.5 cm. Find BC.**

DE=1/2 BC, so BC=2×DE.

BC=2×4.5

BC=9 cm

Final answer: 9 cm.

### Example 4

**In △ABC, D is the midpoint of AB. A line through D parallel to BC meets AC at E. If AC = 14 cm, find AE.**

This uses the converse: a line through one midpoint, parallel to a second side, bisects the third side. So E is the midpoint of AC.

AE=1/2 AC

AE=7 cm

Final answer: 7 cm.

### Example 5

**The sides of △ABC are 12 cm, 16 cm, and 20 cm. Find the perimeter of the triangle formed by joining the midpoints of its three sides.**

Each of the three midsegments is half the side it is parallel to. So the midpoint triangle's sides are:
- 6 cm
- 8 cm
- 10 cm

Perimeter =6+8+10=24 cm.

Final answer: 24 cm — exactly half the original triangle's perimeter of 48 cm.

### Example 6

**In quadrilateral ABCD, P, Q, R, S are the midpoints of sides AB, BC, CD, DA. Show that PQRS is a parallelogram.**

Draw the diagonal AC. In △ABC, P and Q are midpoints of AB and BC, so by the midpoint theorem PQ∥AC and PQ=1/2 AC.

In △ADC, S and R are midpoints of AD and CD, so SR∥AC and SR=1/2 AC.

Therefore PQ∥SR and PQ=SR.

Final answer: One pair of opposite sides of PQRS is equal and parallel, so **PQRS is a parallelogram**. This result, called Varignon's theorem, falls straight out of two applications of the midpoint theorem.

## Why The Midpoint Theorem Matters: Halving Without Measuring

The theorem's power is that it turns a hard-to-reach length into an easy one, and it does so without any measuring tool touching the far side.

- **It computes inaccessible distances.** Surveyors and navigators use midpoint reasoning to find a distance across water or terrain they cannot cross, then double the accessible half.
- **It underpins the coordinate midpoint.** The coordinate midpoint formula is the algebraic face of this same halving idea, letting the theorem run inside coordinate proofs.
- **It scales triangles cleanly.** The midpoint triangle is a half-size copy of the original.

This halving-and-parallel structure is the seed of similar-triangle theory, which is why the midpoint theorem is usually a student's first proof that connects _equal division_ to _parallelism_.

## What Are The Most Common Mistakes With The Midpoint Theorem?

Two mistakes cause most wrong answers, and both come from mishandling the factor of one half.

### Mistake 1: Forgetting the factor of one half

**Where it slips in:** When the midsegment length is given and the third side is wanted, the halving is easy to reverse the wrong way - or drop entirely.

**Don't do this:** Setting the midsegment equal to the third side.

**The correct way:** The midsegment is the _half_; the third side is the _double_. Going from midsegment to third side, multiply by 2.

### Mistake 2: Applying the theorem when the points are not midpoints

**Where it slips in:** When a segment joins two points on the sides that merely _look_ central but are not stated to be midpoints.

**Don't do this:** Using DE=1/2 BC for a segment DE whose endpoints are not confirmed midpoints.

**The correct way:** The theorem needs _both_ endpoints to be true midpoints. If only one is, use the converse instead.

## Conclusion

- The **midpoint theorem** says the segment joining two midpoints of a triangle is parallel to the third side and half its length.
- The proof extends the midsegment to build a parallelogram, then reads off both results from its properties.
- The converse reverses it: a midpoint plus a parallel line locates the second midpoint.
- The segment involved is the triangle's midsegment.

## Practice These To Solidify Your Understanding

Work through these, then check:
1. In △ABC, D and E are midpoints of AB and AC. If BC = 18 cm, find DE. _(Answer to Question 1: 9 cm.)_
2. A midsegment measures 7.5 cm. Find the third side it is parallel to. _(Answer to Question 2: 15 cm.)_
3. A triangle has sides 8, 10, and 14 cm. Find the perimeter of its midpoint triangle. _(Answer to Question 3: 16 cm.)_