Straight Lines cut in Same Ratio by Parallel Planes

Proof

 * Euclid-XI-17.png

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

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

So by :
 * 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 :
 * $AE : EB = AO : OD$

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

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

Therefore from :
 * $AO : OD = CF : FD$

From :
 * $AE : EB = CF : FD$