Construction of Similar Polygon

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


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

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 $AB$.


 * Euclid-VI-18.png

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

On the straight line $BG$ we similarly use Construction of an 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, $CDEF$ is similar $ABHG$.