Straight Lines cut in Same Ratio by Parallel Planes

Theorem

In the words of Euclid:

If two straight lines be cut by parallel planes, they will be cut in the same ratios.

(The Elements: Book $\text{XI}$: Proposition $17$)

Proof

Let the two straight lines $AB$ and $CD$ be cut by the parallel planes $CH$, $KL$ and $MN$ at the points $A, E, B$ and $C, F, D$.

It needs to be demonstrated that:

$AE : EB = CF : FD$

Let $AC$, $BD$ and $AD$ be joined.

Let $AD$ meet $KL$ at $O$.

Let $ED$ and $OF$ be joined.

We have that the parallel planes $KL$ and $MN$ are cut by the plane $EBDO$.

So by Proposition $16$ of Book $\text{XI} $: Common Sections of Parallel Planes with other Plane are Parallel:

their common sections $EO$ and $BD$ are parallel.

Similarly, the parallel planes $GH$ and $KL$ are cut by the plane $AOFC$.

So by Proposition $16$ of Book $\text{XI} $: Common Sections of Parallel Planes with other Plane are Parallel:

their common sections $AC$ and $OF$ are parallel.

We have that the straight line $EO$ is parallel to $BD$.

But $BD$ is one of the sides of $\triangle ABD$.

Therefore from Proposition $2$ of Book $\text{VI} $: Parallel Transversal Theorem:

$AE : EB = AO : OD$

Similarly, the straight line $OF$ is parallel to $AC$.

But $AC$ is one of the sides of $\triangle ADC$.

Therefore from Proposition $2$ of Book $\text{VI} $: Parallel Transversal Theorem:

$AO : OD = CF : FD$

From Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:

$AE : EB = CF : FD$

$\blacksquare$

Historical Note

This proof is Proposition $17$ of Book $\text{XI}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{XI}$. Propositions