Varignon’s Theorem — Midpoints of Any Quadrilateral Form a Parallelogram
Take any lopsided quadrilateral. Join the midpoints of its four sides and you always get a parallelogram — no matter how skewed.
Do not write an equation yet. Shake the figure first.
Drag one vertex and wreck the quadrilateral
Even when the outer shape becomes skewed or concave, its four midpoints rebuild a parallelogram.
Play the reversal slowly
Now move the same pieces step by step and prove that the pattern you touched was not an accident.
An arbitrary quadrilateral; mark the midpoints P, Q, R, S of its sides.
Return to the exact question
Show that joining the midpoints of the four sides of any quadrilateral yields a parallelogram.
Why this approach works
Draw diagonal AC. In triangle ABC, the segment PQ joining the midpoints of AB and BC is the midsegment — parallel to AC and half its length. Likewise SR (midpoints of CD, DA) is parallel to AC and half. So PQ ∥ SR and equal. Using diagonal BD gives QR ∥ PS. Two pairs of parallel equal sides — PQRS is a parallelogram, whatever the original shape. (Its area is exactly half the quadrilateral.)
Proof
PQ and SR are both parallel to diagonal AC and half its length, so PQ∥SR; BD gives QR∥PS. Two pairs of parallel sides make PQRS a parallelogram, independent of the original shape.
Try it yourself
Drag a vertex anywhere (even concave): the midpoint quadrilateral stays a parallelogram.
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