# Construction of Similar Polygon

## Theorem

On any given straight line it is possible to construct a polygon similar to any given polygon.

In the words of Euclid:

*On any given straight line to describe a rectilineal figure similar and similarly situated to a given rectilineal figure.*

(*The Elements*: Book $\text{VI}$: Proposition $18$)

## Construction

Let $AB$ be the given straight line and let $CDEF$ be the given polygon.

Thus we need to construct a polygon similar to $CDEF$ on $AB$.

Join $DF$ and use Construction of Equal Angle to construct $\angle GAB = \angle FCD$ and $\angle ABG = \angle CDF$.

On the straight line $BG$ we similarly use Construction of Equal Angle to construct $\angle BGH = \angle DFE$ and $\angle GBH = \angle FDE$.

Then $ABHG$ is a polygon similar to $CDEF$.

## Proof

From Sum of Angles of Triangle Equals Two Right Angles $\angle CFD = \angle AGB$.

So $\triangle FCD$ is equiangular with $\triangle GAB$.

So from Equiangular Triangles are Similar, $\triangle FCD$ is similar to $\triangle GAB$.

So $FD : GB = FC : GA = CD : AB$.

Similarly from Sum of Angles of Triangle Equals Two Right Angles $\angle GHB = \angle FED$.

So $\triangle FDE$ is equiangular with $\triangle GBH$.

So from Equiangular Triangles are Similar, $\triangle FDE$ is similar to $\triangle GBH$.

So $FD : GB = FE : GH = ED : HB$.

Thus we have that:

- $FC : AG = CD : AB = FE : GH = ED : HB$

Since $\angle CFD = \angle AGB$ and $\angle DFE = \angle BGH$, we have that

- $\angle CFE = \angle CFD + \angle DFE = \angle AGB + \angle BGH = \angle AGH$

For the same reason:

- $\angle CDE = \angle CDF + \angle FDE = \angle ABG + \angle GBH = \angle ABH$

So $CDEF$ is equiangular with $ABHG$.

As has been shown, the sides of these polygons are proportional about their equal angles.

So from Book $\text{VI}$ Definition $1$: Similar Rectilineal Figures, $CDEF$ is similar $ABHG$.

$\blacksquare$

## Historical Note

This proof is Proposition $18$ of Book $\text{VI}$ of Euclid's *The Elements*.

In Proposition $20$ of Book $\text{VI} $: Similar Polygons are composed of Similar Triangles, it is shown by dividing any polygon into triangles, any two similar polygons are composed of similar triangles.

Thus the construction as given here can be seen directly to extend to polygons with any number of sides.

## Sources

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