Construction of Similarly Cut Straight Line

Construction
Let $AB$ be the given uncut straight line.

Let $AC$ be the straight line cut at $D$ and $E$.

Let $AB$ and $AC$ be placed so as to contain any angle.

Join $CB$ and construct $DF$ and $EG$ parallel to $BC$.

Then construct $DHK$ parallel to $AB$.

Then $FG$ is the required cut of $AB$.


 * Euclid-VI-10.png

Proof
We have that each of $FH$ and $HB$ is a parallelogram.

So from Opposite Sides and Angles of Parallelogram are Equal:
 * $DH = FG$ and $HK = GB$

We have that $HE \parallel KC$, and $KC$ is one of the sides of $\triangle DKC$.

So from Parallel Transversal Theorem:
 * $CE : ED = KH : HD$

But $KH = BG$ and $HD = GF$.

So:
 * $CE : ED = BG : GF$

Again, we have $FD \parallel GE$, and $GE$ is one of the sides of $\triangle AGE$.

So from Parallel Transversal Theorem:
 * $ED : DA = GF : FA$

So $AB$ has been cut at $F$ and $G$ similarly to $AC$ at $D$ and $E$.