The Midsegment Theorem
Join two midpoints and you get a parallel segment, half as long.
The explanation
A midsegment joins the midpoints of two sides of a triangle. It always does two things at once:
- it is parallel to the third side, and
- it is exactly half as long.
So if the third side is 14, the midsegment is 7, and they never meet.
Drawing all three midsegments cuts the triangle into four smaller triangles, all congruent to each other and similar to the original with a scale factor of ½.
The theorem is convenient in coordinate geometry: it gives a parallel line and a length in one step, which is often faster than computing slopes and distances separately.
The Triangle Midsegment Theorem states that the segment joining the midpoints of two sides is parallel to the third side and equal to half its length.
The coordinate proof is short: placing the triangle with vertices (0,0), (2a,0) and (2b,2c) makes the midpoints (a,0) and (a+b,c), whose connecting segment has slope c/b, matching the third side, and length √(b² + c²), exactly half.
It follows from, and is a special case of, the Side-Splitter Theorem with the ratio fixed at ½; the parallel conclusion is the converse direction of the proportionality relationship.
The three midsegments form the medial triangle, similar to the original with ratio ½, hence a quarter of the area and half the perimeter. The four resulting small triangles are congruent to one another.
The corresponding result for trapezoids — the midsegment is parallel to both bases and equal to their average — reduces to this one by drawing a diagonal.
Worked example
In △ABC, D and E are midpoints of AB and AC. If DE = 3x − 4 and BC = 4x + 6, find BC.
- Midsegment is half the third side: 2(DE) = BC.
- 2(3x − 4) = 4x + 6.
- 6x − 8 = 4x + 6, so 2x = 14 and x = 7.
- BC = 4(7) + 6.
Answer: BC = 34 (and DE = 17)
Common mistakes
- Setting the midsegment equal to the third side instead of half of it.
- Applying the theorem to a segment joining points that are not both midpoints.