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

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:

and:

it follows that:
 * $\angle AFK = \angle KFB$

Hence from :
 * 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 :
 * $\angle GFB = 2 \cdot \angle BFM$

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 :
 * $\angle AFB = \angle BNF$

Therefore from :
 * $AB : BF = FB : BN$

Therefore from :
 * $AB \cdot BN = BF^2$

We have that:
 * $AL = LK$

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

Therefore from :
 * $KN = AN$

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 :
 * $\angle AKB = \angle KNA$

Therefore:
 * $\triangle AKB$ is equiangular with $\triangle KNA$.

Therefore from :
 * $BA : AK = KA : AN$

Therefore from :
 * $BA \cdot AN = AK^2$

But we have that:
 * $AB \cdot BN = BF^2$

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

But from :
 * $AB \cdot BN + BA \cdot AN = BA^2$

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

Hence the result.