harmonic mean in trapezoid

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

$$\mathrm{\Delta }CDE\sim \mathrm{\Delta }ABE$$

with line ratio  ${\displaystyle \frac{k}{h}}={\displaystyle \frac{CD}{AB}}$, 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

$$\mathrm{\Delta }PED\sim \mathrm{\Delta }ABD$$

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 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 $CD$.