Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ through $AB$ and $AC$.
Extend $DE$ to $DF$ so $DE = EF$.
From Quadrilateral with Bisecting Diagonals is Parallelogram, $ADCF$ is a parallelogram.
By definition of a parallelogram, $AB \parallel CF$.
From Opposite Sides and Angles of Parallelogram are Equal, $AD = CF$.
But $AD = DB$ as $D$ is the midpoint of $AB$.
So $DB = CF$ and $DB \parallel CF$.
Thus also by Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel $DF = BC$ and $DF \parallel BC$.
As $DE = EF$, $DE$ is the midpoint of $DF$ and so $DE = \dfrac 1 2 DF$.
Thus $DE = \dfrac 1 2 BC$ and $DE \parallel BC$.
Hence the result.
Also known as
The midline theorem is also known as the midpoint theorem or mid-point theorem.
- 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.26$: Corollary $1$
- 1989: Ephraim J. Borowski and Jonathan M. Borwein: Dictionary of Mathematics ... (previous) ... (next): Entry: midpoint theorem
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): Entry: mid-point theorem
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Entry: midpoint theorem