Problem — Thales' theorem on a pencil of lines cut by parallels

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Two lines issuing from the same point $V$ are cut by two parallel lines. State and prove Thales’ theorem: the segments cut off on the two lines are proportional. Then verify the proportion on a numerical instance.

Solution

Statement. From the point $V$ two lines issue; a first parallel meets them at $A$ and $B$, a second parallel (on the same side) meets them at $C$ and $D$ (with $A,C$ on the first line and $B,D$ on the second). Then $\frac{\overline{VA}}{\overline{AC}}=\frac{\overline{VB}}{\overline{BD}}.$

Proof. Consider the triangles $VAB$ and $VCD$. They share the angle $\hat V$; moreover, since $AB\parallel CD$, the angle $\widehat{VAB}\cong\widehat{VCD}$ (corresponding angles). By the first similarity criterion $VAB\sim VCD$, so the homologous sides are proportional: $\frac{\overline{VA}}{\overline{VC}}=\frac{\overline{VB}}{\overline{VD}}.$ Applying the separation property to the two equal fractions (subtracting $1$ from each side after writing $\overline{VC}=\overline{VA}+\overline{AC}$ and $\overline{VD}=\overline{VB}+\overline{BD}$) yields equivalently $\frac{\overline{VA}}{\overline{AC}}=\frac{\overline{VB}}{\overline{BD}}.\qquad\blacksquare$

Numerical check. Let $\overline{VA}=6$, $\overline{AC}=9$ and $\overline{VB}=8$; compute $\overline{BD}$: $\frac{\overline{VA}}{\overline{AC}}=\frac{\overline{VB}}{\overline{BD}}\ \Longrightarrow\ \overline{BD}=\frac{\overline{VB}\cdot\overline{AC}}{\overline{VA}}=\frac{8\cdot 9}{6}=12.$ Check: $\dfrac{\overline{VA}}{\overline{AC}}=\dfrac{6}{9}=\dfrac{2}{3}$ and $\dfrac{\overline{VB}}{\overline{BD}}=\dfrac{8}{12}=\dfrac{2}{3}$: the proportion holds. $\boxed{\overline{BD}=12.}$