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Worked solution

Prove the triangle midpoint theorem with vectors

Show that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half its length.

CJC · Vectors Basic · Example 2

Answer

MN→=12BC→\overrightarrow{MN}=\tfrac12\overrightarrow{BC}, so MN∥BCMN\parallel BC and MN=12BCMN=\tfrac12BC.

Compare your reasoning

Name the source midpoints

MA→=12BA→,AN→=12AC→\overrightarrow{MA}=\frac12\overrightarrow{BA},\qquad\overrightarrow{AN}=\frac12\overrightarrow{AC}

Let M and N be the midpoints of AB and AC respectively in the nondegenerate triangle ABC.

Follow the path through A

MN→=MA→+AN→=12BA→+12AC→=12(BA→+AC→)=12BC→.\begin{aligned}\overrightarrow{MN}&=\overrightarrow{MA}+\overrightarrow{AN}\\&=\frac12\overrightarrow{BA}+\frac12\overrightarrow{AC}\\&=\frac12(\overrightarrow{BA}+\overrightarrow{AC})\\&=\frac12\overrightarrow{BC}.\end{aligned}

The triangle law and distributive law give every stage of the source proof.

Read direction and length separately

MN∥BC,∣MN→∣=12∣BC→∣⟹MN=12BC.MN\parallel BC,\qquad\lvert\overrightarrow{MN}\rvert=\frac12\lvert\overrightarrow{BC}\rvert\quad\Longrightarrow\quad MN=\frac12BC.

The positive multiplier gives the same direction; taking moduli proves the length statement. The side BC is nonzero.

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Triangle ABC has M halfway along AB and N halfway along AC, matching the Teacher completion. MN is parallel to BC with half its length. No numerical side lengths are imposed.
Triangle ABC has M halfway along AB and N halfway along AC, matching the Teacher completion. MN is parallel to BC with half its length. No numerical side lengths are imposed.
Triangle ABC has M halfway along AB and N halfway along AC, matching the Teacher completion. MN is parallel to BC with half its length. No numerical side lengths are imposed.