Distance between Two Parallel Straight Lines is Everywhere the Same

Theorem

Let $AB$ and $CD$ be parallel straight lines.

Let perpendiculars $EF$ and $GH$ be drawn from $AB$ to $CD$, where $E, G$ are on $AB$ and $F, H$ are on $CD$.

Then $EF = GH$.

That is, the distance between $AB$ and $CD$ is the same everywhere along their length.

Proof

$\angle EFH, \angle FEG, \angle EGH, \angle FHG$ are all right angles.

Then $EF$ and $GH$ are parallel.

Thus $\Box EFHG$ is by definition a parallelogram.

From Opposite Sides and Angles of Parallelogram are Equal it follows that $EF = GH$.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.22$: Corollary $1$