Similar Polygons are composed of Similar Triangles

Theorem

 * Similar polygons are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes, and the polygon has to the polygon a ratio duplicate of that which the corresponding side has to corresponding side.

Proof
Let $ABCDE$ and $FGHKL$ be similar polygons, and let $AB$ correspond to $FG$.

We need to show that $ABCDE$ and $FGHKL$ are divided into similar triangles, and into triangles equal in multitude and in the same ratio as the wholes.

Also that the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$.


 * Euclid-VI-20.png

Join up $BE, EC, GL, LH$.

Since $ABCDE$ and $FGHKL$ are similar:
 * $\angle BAE = \angle GFL$

From :
 * $BA : AE = GF : FL$

Thus from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle ABE$ is similar to $\triangle FGL$.

So $\angle ABE = \angle FGL$.

But $\angle ABC = \angle FGH$ because $ABCDE$ and $FGHKL$ are similar.

So:
 * $\angle EBC = \angle LGH$

Because $\triangle ABE$ is similar to $\triangle FGL$:
 * $EB : BA = LG : GF$

Also, because $ABCDE$ and $FGHKL$ are similar:
 * $AB : BC = FG : GH$

So from Equality of Ratios Ex Aequali:
 * $EB : BC = LG : GH$

So from Triangles with One Equal Angle and Two Sides Proportional are Similar, $\triangle EBC$ is similar to $\triangle LGH$.

For the same reason, $\triangle ECD$ is similar to $\triangle LHK$.

So $ABCDE$ and $FGHKL$ have been divided into similar triangles, and into triangles equal in multitude.

Now let $AC, FH$ be joined.

Because $ABCDE$ and $FGHKL$ are similar:
 * $\angle ABC = \angle FGH$

From Triangles with One Equal Angle and Two Sides Proportional are Similar $\triangle ABC$ is similar to $\triangle FGH$.

Therefore $\angle BAC = \angle GFH$ and $\angle BCA = \angle GHF$.

Also, we have that $\angle BAM = \angle GFN$, and $\angle ABM = \angle FGN$.

So from Sum of Angles of Triangle Equals Two Right Angles $\angle AMB = \angle FGN$ and so $\triangle ABM$ is similar to $\triangle FGN$.

Similarly we can show that $\triangle BMC$ is similar to $\triangle GNH$.

Therefore $AM : MB = FN : NG$ and $BM : MC = FN : NH$.

But from Areas of Triangles and Parallelograms Proportional to Base:
 * $AM : MC = \triangle ABM : \triangle MBC$.

So from Sum of Components of Equal Ratios:
 * $\triangle ABM : \triangle MBC = \triangle ABE : \triangle CBE$

But:
 * $\triangle ABM : \triangle MBC = AM : MC$

So:
 * $\triangle ABE : \triangle CBE = AM : MC$

For the same reason:
 * $FN : NH = \triangle FGL : \triangle GLH$

As $AM : MC = FN : NH$, it follows that:
 * $\triangle ABE : \triangle BEC = \triangle FGL : \triangle GLH$

Similarly:
 * $\triangle ABE : \triangle FGL = \triangle BEC : \triangle GLH$

We now join $BD$ and $GK$, and by a similar construction show that:
 * $\triangle BEC : \triangle LGH = \triangle ECD : \triangle LHK$

From Sum of Components of Equal Ratios:
 * $ABE : FGL = ABCDE : FGHKL$

But from Ratio of Areas of Similar Triangles $\triangle ABE$ has to $\triangle FGL$ a ratio duplicate of $AB : FG$.

Therefore the area of the polygon $ABCDE$ has to the polygon $FGHKL$ a ratio duplicate of $AB : FG$.