Thales' theorem, equidecomposability and triangle-parallelogram

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• (a) State the (great) Thales’ theorem.
• (b) State the concept of equidecomposability.
• (c) Prove that a triangle is equidecomposable with a parallelogram having the same base and height equal to half that of the triangle.

Solution

(a) Thales’ theorem (intercept theorem). A pencil of parallel lines, cut by two transversals, determines on them proportional segments: if $a\parallel b\parallel c$ cut the transversals $r$ and $s$ at $A,B,C$ and $A',B',C'$, then $\frac{\overline{AB}}{\overline{BC}}=\frac{\overline{A'B'}}{\overline{B'C'}}.$

(b) Equidecomposability. Two plane figures are equidecomposable if they can be split into the same finite number of polygonal pieces, pairwise congruent. Equidecomposable figures are in particular equivalent (same area).

(c) Triangle and parallelogram. Let $ABC$ be the triangle, $AB$ the base, $h$ the height from $C$. Let $M,N$ be the midpoints of $CA,CB$; by the (small) Thales’ theorem $MN\parallel AB$ and $MN$ is at distance $\dfrac{h}{2}$ from $AB$. Cut the triangle along $MN$: obtain the small top triangle $CMN$ and the trapezoid $ABNM$. Rotate $CMN$ by $180^\circ$ about the midpoint $N$ (of $CB$): $C$ goes to $B$, and the small triangle fits beside the trapezoid forming a parallelogram of base $AB$ and height $\dfrac{h}{2}$. The triangle and the parallelogram are made of the same two pieces: hence they are equidecomposable. $\boxed{\text{Triangle}\ \equiv\ \text{parallelogram of equal base and height }\tfrac h2\ \Rightarrow\ \mathcal A_\triangle=\tfrac12\,b\,h.}$