Midpoint theorem (triangle)

The midpoint theorem, midsegment theorem, or midline theorem states that if the midpoints of two sides of a triangle are connected, then the resulting line segment will be parallel to the third side and have half of its length. The midpoint theorem generalizes to the intercept theorem, where rather than using midpoints, both sides are partitioned in the same ratio.^[1][2]

${\displaystyle {\begin{aligned}&{\text{D and E midpoints of AC and BC}}\\\Rightarrow \,&DE\parallel AB{\text{ and }}2|DE|=|AB|\end{aligned}}}$

The converse of the theorem is true as well. That is if a line is drawn through the midpoint of triangle side parallel to another triangle side then the line will bisect the third side of the triangle.

The triangle formed by the three parallel lines through the three midpoints of sides of a triangle is called its medial triangle.

Proof

Proof by construction

Proof

Given: In a ${\displaystyle \triangle ABC}$ the points M and N are the midpoints of the sides AB and AC respectively.

Construction: MN is extended to D where MN=DN, join C to D.

To Prove:

• ${\displaystyle MN\parallel BC}$
• ${\displaystyle MN={1 \over 2}BC}$

Proof:

• ${\displaystyle AN=CN}$ (given)
• ${\displaystyle \angle ANM=\angle CND}$ (vertically opposite angle)
• ${\displaystyle MN=DN}$ (constructible)

Hence by Side angle side.

${\displaystyle \triangle AMN\cong \triangle CDN}$

Therefore, the corresponding sides and angles of congruent triangles are equal

• ${\displaystyle AM=BM=CD}$
• ${\displaystyle \angle MAN=\angle DCN}$

Transversal AC intersects the lines AB and CD and alternate angles ∠MAN and ∠DCN are equal. Therefore

• ${\displaystyle AM\parallel CD\parallel BM}$

Hence BCDM is a parallelogram. BC and DM are also equal and parallel.

• ${\displaystyle MN\parallel BC}$
• ${\displaystyle MN={1 \over 2}MD={1 \over 2}BC}$,

Q.E.D.

Proof by similar triangles

Proof

Let D and E be the midpoints of AC and BC.

To prove:

• ${\displaystyle DE\parallel AB}$,
• ${\displaystyle DE={\frac {1}{2}}AB}$.

Proof:

${\displaystyle \angle C}$ is the common angle of ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle DEC}$.

Since DE connects the midpoints of AC and BC, ${\displaystyle AD=DC}$, ${\displaystyle BE=EC}$ and ${\displaystyle {\frac {AC}{DC}}={\frac {BC}{EC}}=2.}$ As such, ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle DEC}$ are similar by the SAS criterion.

Therefore, ${\displaystyle {\frac {AB}{DE}}={\frac {AC}{DC}}={\frac {BC}{EC}}=2,}$ which means that ${\displaystyle DE={\frac {1}{2}}AB.}$

Since ${\displaystyle \triangle ABC}$ and ${\displaystyle \triangle DEC}$ are similar and ${\displaystyle \triangle DEC\in \triangle ABC}$, ${\displaystyle \angle CDE=\angle CAB}$, which means that ${\displaystyle AB\parallel DE}$.

Q.E.D.

See also

• Median of the trapezoid theorem