Equidecomposability of a triangle and a parallelogram

A first example of equidecomposability relates two fundamental figures: the triangle and the parallelogram.

Theorem

A triangle is equidecomposable with a parallelogram having the same base as its base and half the triangle’s height as its height.

Proof

From the midpoint $M$ of side $AC$ the parallelogram $AMDB$ is constructed, equidecomposable with the triangle $ABC$.

1. From the midpoint $M$ of $AC$ draw the parallel to $AB$.
2. From $B$ draw the parallel to $AC$, which meets the previous line at $D$.
3. The upper triangle $CME$ “flips over” exactly into the space $BED$: the two figures are congruent.
4. The result is the parallelogram $AMDB$, with base $AB$ and height $=\tfrac{1}{2}CH$.

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Links

Topics: Equivalence and Pythagoras
Concepts: Area · Equidecomposability · Equivalence · Parallelogram · Triangle
Methods: Equivalence of figures
Skills: Proving · Synthetic geometry