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Midsegment of a Triangle

Midsegment of a Triangle – definition and example, Geometry math glossary – Mathovore

Key takeaway: A midsegment of a triangle is a segment that links the midpoints of two sides; it runs parallel to the third side and measures exactly half of it.

Grade 10 – GeometryG-CO.C.10Geometry and Measurement

Definition of Midsegment of a Triangle

Definition

The midsegment of a triangle connects the centers of two of its sides. Every triangle has three of them, and together they cut it into four small triangles that are all congruent.

The midsegment theorem gives two facts at once. It lies parallel to the side it never touches, and it measures half of that side’s length. These facts make it a quick tool for finding unknown lengths and for proving lines parallel.

Formula and notation

For the midsegment of a triangle \(\overline{DE}\), with \(D\) halfway along side \(AB\) and \(E\) halfway along side \(AC\):

\[ \overline{DE} \parallel \overline{BC} \quad \text{and} \quad DE = \tfrac{1}{2}\, BC \]

The third side \(BC\) is called the base for this relationship.

ABCDEDE = 17BC = 34
Midpoints D and E are joined; DE is parallel to BC and half as long.

Key facts

  • The midsegment is never longer than half of the side it faces.
  • The three midsegments form a medial triangle with half the perimeter.
  • Doubling the midsegment gives the length of the parallel side.
  • The triangle formed by the midsegment and the top vertex is similar to the whole triangle.

Midsegment of a Triangle example

Problem

A roof truss is a triangle with a brace \(DE\) joining the centers of two rafters, parallel to the floor beam \(BC\). The brace measures \(3x – 4\) inches and the beam \(5x – 1\) inches. Find \(x\) and both lengths.

  1. The midsegment is half the base: \(3x – 4 = \frac{1}{2}(5x – 1)\).

  2. Multiply by 2: \(6x – 8 = 5x – 1\), so \(x = 7\).

  3. Then \(DE = 3(7) – 4 = 17\) and \(BC = 5(7) – 1 = 34\).

\(x = 7\), \(DE = \) 17 and \(BC = \) 34.

Common mistake

Watch out

Cal sets the midsegment equal to the base, writing \(3x – 4 = 5x – 1\), and gets a negative \(x\).

Why it is wrong

The brace has only half the length of the beam it runs alongside; it is not equal to it. The equation must be \(3x – 4 = \frac{1}{2}(5x – 1)\), or equivalently \(2(3x – 4) = 5x – 1\).

Related terms and lessons

Browse the glossary: A–Z list · Geometry vocabulary · Geometry and Measurement

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