Construction of Similarly Cut Straight Line
Theorem
In the words of Euclid:
- To cut a given uncut straight line similarly to a given cut straight line.
(The Elements: Book $\text{VI}$: Proposition $10$)
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 $AB$.
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 $E$.
$\blacksquare$
Historical Note
This proof is Proposition $10$ of Book $\text{VI}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{VI}$. Propositions