Join two midpoints and you get a shrunken, parallel copy of the third side.
Here is where the parallelogram pays a debt back to the triangle. Mark the midpoints of two sides of any triangle and join them with a straight segment. That little segment — a midsegment — turns out to be a miniature of the third side: it runs exactly parallel to it and measures exactly half as long, no matter how the triangle is shaped. One clever parallelogram, bolted onto the triangle, makes the whole thing obvious. And once you have it, the midsegment becomes a workhorse — a fast way to find an unknown length and a clean way to prove that two lines are parallel.
Take any triangle △ABC. Let D be the midpoint of side AB and E the midpoint of side AC. The segment DE that joins them is called a midsegment of the triangle (some books call it a midline).
A triangle has three sides, so you can choose midpoints two at a time in three ways — every triangle has three midsegments, one parallel to each side.
A midsegment joins the midpoints of two sides — both endpoints sit on the edges of the triangle. A median runs from a vertex to the midpoint of the opposite side. Different endpoints, different segment, different job. In the figure below, DE is a midsegment; the faint slate segment from A is a median.
Now for the result that makes the midsegment worth knowing. It says two things at once — a direction and a length.
The segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
DE ∥ BC and DE = ½ BC
Why is this true? We borrow the parallelogram we just studied. The trick is to double the midsegment and watch a parallelogram appear.
Extend the midsegment DE past E to a new point F, choosing F so that EF = DE. Then draw CF. Watch the chain of reasoning:
| Statement | Reason |
|---|---|
| AE = EC | E is the midpoint of AC |
| DE = EF | we chose F that way |
| ∠AED = ∠CEF | vertical angles at E |
| △ADE ≅ △CFE | SAS |
| CF = AD = DB, CF ∥ AB | congruent parts; AD = DB since D is a midpoint |
| DBCF is a parallelogram | one pair of sides (DB, CF) equal & parallel |
| DF ∥ BC and DF = BC | opposite sides of ▱DBCF |
| DE ∥ BC and DE = ½ BC | DE is half of DF |
The key move is the test from Stage 16.3: one pair of sides both parallel and equal makes a parallelogram. Here DB and CF are that pair. Once DBCF is a parallelogram, its other pair of opposite sides, DF and BC, are equal and parallel — and since E is the midpoint of DF, the midsegment DE is exactly half of BC. ∴ DE ∥ BC and DE = ½ BC.
Because DE = ½ BC, the same equation read backward says BC = 2·DE. A midsegment is half its side; a side is twice its midsegment. Knowing either one hands you the other.
Join all three midpoints of a triangle and something tidy happens: the triangle splits into four smaller triangles, and all four are congruent to one another. Each has the same shape as △ABC with every side half as long — you'll meet this idea as similarity in Stage 17.
The central triangle △DEF — the one built entirely from midsegments — has a special name: the medial triangle. Since each of its sides is half a side of △ABC, its perimeter is half the perimeter of the original.
In △ABC, D and E are the midpoints of AB and AC.
• If BC = 10, then DE = ½ · 10 = 5.
• If the midsegment DE = 7, then the side BC = 2 · 7 = 14.
• Because DE ∥ BC for free, the midsegment is also a quick way to prove two lines parallel: show a segment joins two midpoints, and you may conclude it is parallel to the third side at once.
• A midsegment joins the midpoints of two sides of a triangle. (A median goes from a vertex to a midpoint — not the same thing.)
• Every triangle has exactly three midsegments.
• Midsegment theorem: a midsegment is parallel to the third side and half its length — DE ∥ BC, DE = ½ BC.
• Read backward: a side is twice its midsegment, BC = 2·DE.
• Joining all three midpoints makes four congruent triangles; the central medial triangle has half the perimeter of the original.
• The proof works by doubling the midsegment to build a parallelogram — the triangle's gift back from Stage 16.2–16.3.
D and E are the midpoints of AB and AC in △ABC, and BC = 12. Find DE.
A midsegment is half the third side, so DE = ½ · 12 = 6.
A midsegment of a triangle measures 9. How long is the side it is parallel to?
The side is twice the midsegment: 2 · 9 = 18.
How many midsegments does a triangle have, and how many medians?
Three midsegments (one for each pair of sides) and three medians (one from each vertex). They are different segments — a midsegment joins two midpoints; a median joins a vertex to a midpoint.
You draw all three midsegments of a triangle. How many small triangles do you create, and how are they related?
Four small triangles, and all four are congruent — each has the same shape as the original with sides half as long. The middle one is the medial triangle.
In △ABC, D and E are midpoints of AB and AC, and DE = (3x − 1) while BC = 22. Solve for x.
By the theorem DE = ½ BC = ½ · 22 = 11. So 3x − 1 = 11 ⇒ 3x = 12 ⇒ x = 4.
△ABC has perimeter 30. Find the perimeter of its medial triangle. Why?
15. Each side of the medial triangle is a midsegment — half of a side of △ABC — so its perimeter is exactly half: ½ · 30 = 15.
Six questions to lock it in. Tap the answer you think is right.
The triangle midsegment is a small theorem with an outsized payoff: it converts the parallelogram facts of the previous lessons into a tool that finds lengths and certifies parallels. The proof is worth lingering over because it models a powerful habit — construct an auxiliary figure. Doubling the midsegment to length DF and recognizing DBCF as a parallelogram (via "one pair of sides equal and parallel") is the same move students will reuse all through geometry. Common Core touchpoints: HS G-CO.C.10 (prove theorems about triangles, including the midsegment), G-SRT.B.5 (use congruence and similarity criteria to solve problems), and informally 8.G.
Two slips are common. First, students conflate the midsegment with the median — both involve a midpoint, but a median touches a vertex while a midsegment never does. Second, they remember "midsegment relates to the third side" but forget the half, writing DE = BC instead of DE = ½ BC. The interactive figure is the antidote: it shows the midsegment visibly fitting twice into the base for every shape of triangle. Ask "if BC = 10, how long is DE?" and listen for "5," not "10."