# Thales' Theorem

A theorem that is credited to Thales (after Thales of Miletus, (c. 620 - c. 546 BC) is the subject of Euclid's Elements VI.2:

If a straight line is drawn parallel to one of the sides of a triangle, then it cuts the other sides of the triangle, or these produced, proportionally; and, if the sides of the triangle, or the sides produced, are cut proportionally, then the line joining the points of section is parallel to the remaining side of the triangle.

(The applet below illustrates the theorem and its proof. Points A, B, C, D are draggable.)

24 December 2013, Created with GeoGebra

### Proof

Given DE||BC, we want to prove that BD/AD = CE/AE.

Join BE, CD. Then triangles BDE and CDE have equal areas by Elements I.38: Area(BDE) = Area(CDE). Triangles ADE and BDE share an altitude (from E) and, therefore, by Elements VI.1, Area(ADE)/Area(BDE) = AD/BD.

Turning to triangles CDE and ADE, we similarly observe that Area(ADE)/Area(CDE) = AE/CE. But, since Area(BDE) = Area(CDE), we also have Area(ADE)/Area(BDE) = Area(ADE)/Area(CDE), implying

as required.

Now for the converse. Given AD/BD = AE/CE, prove that DE||BC.

For a proof, note that the steps above are reversible. First we have Area(ADE)/Area(BDE) = AD/BD and Area(ADE)/Area(CDE) = AC/CE, implying Area(ADE)/Area(BDE) = Area(ADE)/Area(CDE). I.e., Area(BDE) = Area(CDE), while the two triangle share the base DE. By Elements I.39, DE||BC. Note that AD/BD = AC/CE is equivalent to AB/BD = AC/CE. In particular, triangles ABC and ADE are similar so that their bases are in the same ratio:

AB/BD = AC/CE = BC/DE.

As a consequence, if DE is a midline of ΔABC, i.e., if AD = BD and AE = CE, then DE is half BC: DE = BC/2. • 