Square on Side of Regular Pentagon inscribed in Circle equals Squares on Sides of Hexagon and Decagon inscribed in same Circle

From ProofWiki
Jump to navigation Jump to search

Theorem

In the words of Euclid:

If an equilateral pentagon be inscribed in a circle, the square on the side of the pentagon is equal to the squares on the side of the hexagon and on that of the decagon inscribed in the same circle.

(The Elements: Book $\text{XIII}$: Proposition $10$)


Proof

Euclid-XIII-10.png

Let $ABCDE$ be a circle.

Let the regular pentagon $ABCDE$ be inscribed within the circle $ABCDE$.

It is to be demonstrated that the square on the side of the pentagon $ABCDE$ equals the square on the side of a regular hexagon plus the square on the side of a regular decagon which have both been inscribed within the circle $ABCDE$.


Let the center of the circle $ABCDE$ be $F$.

Let $AF$ be joined and produced to $G$.

Let $FB$ be joined.

Let $FH$ be drawn from $F$ perpendicular to $AB$ and produced to $K$.

Let $AK$ and $KB$ be joined.

Let $FL$ be drawn from $F$ perpendicular to $AK$ and produced to $M$.

Let $KN$ be joined.

We have that the arc $ABCG$ equals the arc $AEDG$.

Within those arcs, the arc $ABC$ equals the arc $AED$.

Therefore the remainders are equal: the arc $CG$ equals the arc $GD$.

But $CD$ is the side of a regular pentagon.

Therefore $CG$ (when joined) is the side of a regular decagon.

We have that:

$FA = FB$

and:

$FH \perp AB$

Therefore from:

Proposition $5$ of Book $\text{I} $: Isosceles Triangle has Two Equal Angles

and:

Proposition $26$ of Book $\text{I} $: Triangle Side-Angle-Angle Congruence:

it follows that:

$\angle AFK = \angle KFB$

Hence from Proposition $26$ of Book $\text{III} $: Equal Angles in Equal Circles:

the arc $AK$ equals the arc $KB$.

Therefore:

the arc $AB$ equals double the arc $BK$.

Therefore $AK$ is the side of a regular decagon.

For the same reason:

the arc $AK$ equals double the arc $KM$.

We have that:

the arc $AB$ equals double the arc $BK$

while:

the arc $CD$ equals the arc $KB$.

Therefore:

the arc $CD$ equals double the arc $BK$.

But:

the arc $AK$ equals double the arc $KM$.

Therefore:

the arc $BK$ equals double the arc $KM$

and therefore:

the arc $CG$ equals double the arc $KM$.

But we have that:

the arc $CB$ equals the arc $BA$

therefore

the arc $CB$ equals double the arc $BK$.

Therefore:

the arc $GB$ equals double the arc $BM$.

So from Proposition $33$ of Book $\text{VI} $: Angles in Circles have Same Ratio as Arcs:

$\angle GFB = 2 \cdot \angle BFM$

But we also have:

$\angle FAB = \angle ABF$

and so:

$\angle GFB = 2 \cdot \angle FAB$

Therefore:

$\angle BFN = \angle FAB$

But $\angle ABF$ is common between $\triangle ABF$ and $\triangle BFN$.

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

$\angle AFB = \angle BNF$

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

$AB : BF = FB : BN$

Therefore from Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:

$AB \cdot BN = BF^2$


We have that:

$AL = LK$

while $LN$ is common and perpenducular to $AK$.

Therefore from Proposition $4$ of Book $\text{I} $: Triangle Side-Angle-Side Congruence:

$KN = AN$

Therefore:

$\angle LKN = \angle LAN$

But:

$\angle LAN = \angle KBN$

Therefore:

$\angle LAN = \angle KBN$

We have that:

$\angle A$ is common to $\triangle AKB$ and $\triangle AKN$.

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

$\angle AKB = \angle KNA$

Therefore:

$\triangle AKB$ is equiangular with $\triangle KNA$.

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

$BA : AK = KA : AN$

Therefore from Proposition $17$ of Book $\text{VI} $: Rectangles Contained by Three Proportional Straight Lines:

$BA \cdot AN = AK^2$

But we have that:

$AB \cdot BN = BF^2$

Therefore:

$AB \cdot BN + BA \cdot AN = BF^2 + AK^2$

But from Proposition $2$ of Book $\text{II} $: Square is Sum of Two Rectangles:

$AB \cdot BN + BA \cdot AN = BA^2$

That is:

$BA^2 = BF^2 + AK^2$

But:

$BA$ is the side of the pentagon $ABCDE$
$AK$ is the side of a regular decagon inscribed within the circle $ABCDE$.

and from Porism to Proposition $15$ of Book $\text{IV} $: Inscribing Regular Hexagon in Circle:

$BF$ is the side of a regular hexagon inscribed within the circle $ABCDE$.

Hence the result.

$\blacksquare$


Historical Note

This proof is Proposition $10$ of Book $\text{XIII}$ of Euclid's The Elements.


Sources