Theorem. If a line parallel to the bases of a trapezoid passes through the intersecting point of the diagonals, then the portion of the line inside the trapezoid is the harmonic mean of the bases.

Proof. Let $AB$ and $DC$ be the bases of a trapezoid $ABCD$ and $E$ the intersecting point of the diagonals of $ABCD$ . Denote the cutting point of $AD$ and the line through $E$ and parallel to the bases by $P$ , and the cutting point of $BC$ and the same line by $Q$ . Then we have $$\Delta CDE \;\sim\; \Delta ABE$$ with line ratio $\displaystyle\frac{k}{h} = \frac{CD}{AB}$ , where $h$ and $k$ are the heights of the triangles $ABE$ and $CDE$ , respectively, when $h\!+\!k$ equals the height of the trapezoid. We have also $$\Delta PED \;\sim\; \Delta ABD$$ with line ratio $$PE:AB \;=\; \frac{k}{h\!+\!k} \;=\; \frac{\frac{k}{h}}{1\!+\!\frac{k}{h}} \;=\; \frac{\frac{CD}{AB}}{1+\frac{CD}{AB}}.$$ Thus we can express the length of $PE$ as $$PE \;=\; AB\cdot\frac{\frac{CD}{AB}}{1+\frac{CD}{AB}} \;=\; \frac{CD}{1+\frac{CD}{AB}} \;=\; \frac{AB\!\cdot\!CD}{AB\!+\!CD}.$$ Similarly we may determine $EQ$ and state that $EQ = PE$ . Consequently, $$PQ \;=\; PE\!+\!EQ \;=\; \frac{2\!\cdot\!AB\!\cdot\!CD}{AB\!+\!CD},$$ which is the harmonic mean of the bases $AB$ and $CD$ .