TheoremProved

Thales's intercept theorem

Statement

In a triangle $ABC$ , let a line intersect the sides $AB$ and $AC$ (or their extensions) at $D$ and $E$ respectively. The line $DE$ is parallel to $BC$ if and only if it divides the sides proportionally: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , and in that case $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC}$ .

Why is it true?

Sliding a line parallel to the base $BC$ toward the apex $A$ scales the triangle $\triangle ADE$ uniformly relative to $\triangle ABC$ , because the angles at the base remain equal to the corresponding angles of $\triangle ABC$ . A uniform rescaling shrinks every linear dimension by the same factor, so the segments cut on $AB$ and on $AC$ , as well as the parallel segment $DE$ itself, all keep the same ratio.

Proof sketch

Follow Euclid's area argument (Book VI, Proposition 2): join $BE$ and $CD$ . Triangles $\triangle ADE$ and $\triangle BDE$ share the altitude from $E$ to the line $AB$ , so the ratio of their areas equals the ratio of their bases, $\dfrac{[\triangle ADE]}{[\triangle BDE]} = \dfrac{AD}{DB}$ ; similarly $\dfrac{[\triangle ADE]}{[\triangle CDE]} = \dfrac{AE}{EC}$ . Because $DE$ is parallel to $BC$ , triangles $\triangle BDE$ and $\triangle CDE$ share base $DE$ and have equal altitudes between the parallel lines, so $[\triangle BDE] = [\triangle CDE]$ , which forces $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ .

Proved by

proved -300Euclid of Alexandria

Topics that use this theorem

Thales' theorem and the Pythagorean theorem

Related theorems

Pythagorean theorem Euclid's parallel postulate

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Euclid (translated by Thomas L. Heath) (1956). The Thirteen Books of Euclid's Elements, Vol. 2 (Book VI, Proposition 2)
Robin Hartshorne (2000). Geometry: Euclid and Beyond · DOI:10.1007/978-0-387-22676-7