# Similar Polygons are composed of Similar Triangles

## Theorem

In the words of Euclid:

*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.*

(*The Elements*: Book $\text{VI}$: Proposition $20$)

## Proof

Let $ABCDE$ and $FGHKL$ be similar polygons such that $AB$ corresponds 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 $AB : FG$.

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

Since $ABCDE$ and $FGHKL$ are similar:

- $\angle BAE = \angle GFL$

From Book $\text{VI}$ Definition $1$: Similar Rectilineal Figures:

- $BA : AE = GF : FL$

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 $AB : FG$.

$\blacksquare$

### Porism

In the words of Euclid:

*Similarly also it can be proved in the case of quadrilaterals that they are in the duplicate ratio of the corresponding sides. And it was also proved in the case of triangles; therefore also, generally, similar rectilineal figures are to one another in the duplicate ratio of the corresponding sides.*

(*The Elements*: Book $\text{VI}$: Proposition $20$ : Porism)

## Historical Note

This proof is Proposition $20$ 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