Areas of Similar Polygons Inscribed in Circles are as Squares on Diameters

From ProofWiki
Jump to navigation Jump to search


In the words of Euclid:

Similar polygons inscribed in circles are to one another as the squares on the diameters.

(The Elements: Book $\text{XII}$: Proposition $1$)



Let $ABC$ and $FGH$ be circles.

Let $ABCDE$ and $FGHKL$ be similar polygons inscribed in $ABC$ and $FGH$ respectively.

Let $BM$ and $GN$ be diameters of $ABC$ and $FGH$ respectively.

It is to be demonstrated that the ratio of the square on $BM$ to the square on $GN$ equals the ratio of the area of $ABCDE$ to the area of $FGHKL$.

Let $BE, AM, GL, FN$ be joined.

We have that $ABCDE$ is similar to $FGHKL$.


$\angle BAE = \angle GFL$

and so from Book $\text{VI}$ Definition $1$: Similar Rectilineal Figures:

$BA : AE = GF : FL$

Thus from Proposition $6$ of Book $\text{VI} $: Triangles with One Equal Angle and Two Sides Proportional are Similar:

$\triangle ABE$ is equiangular with $\triangle GFL$.


$\angle AEB = \angle FLG$

But from Proposition $27$ of Book $\text{III} $: Angles on Equal Arcs are Equal:

$\angle AEB = \angle AMB$


$\angle FLG = \angle FNG$


$\angle AMB = \angle FNG$

But from Proposition $31$ of Book $\text{III} $: Relative Sizes of Angles in Segments:

the right angle $\angle BAM$ equals the right angle $\angle GFN$.

Therefore from Proposition $32$ of Book $\text{I} $: Sum of Angles of Triangle equals Two Right Angles:

$\angle ABM = \angle FGN$

Therefore $\triangle ABM$ is equiangular with $\triangle FGN$.

Therefore from Proposition $4$ of Book $\text{VI} $: Equiangular Triangles are Similar:

$BM : GN = BA : GF$

But the ratio of the square on $BM$ to the square on $GN$ is the duplicate ratio of $BM$ to $GN$.

From Proposition $20$ of Book $\text{VI} $: Similar Polygons are composed of Similar Triangles:

$ABCDE : FGHKL = BA^2 : GF^2$


$ABCDE : FGHKL = BM^2 : GN^2$

Hence the result.


Historical Note

This proof is Proposition $1$ of Book $\text{XII}$ of Euclid's The Elements.