If an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.

Let *ABCDE* be a circle, and let the equilateral pentagon *ABCDE* be inscribed in the circle *ABCDE.*

I say that the square on the side of the pentagon *ABCDE* equals the sum of the squares on the side of the hexagon and on that of the decagon inscribed in the circle *ABCDE.*

Take the center *F* of the circle, join *AF* and carry it through to the point *G,* and join *FB.* Draw *FH* from *F* perpendicular to *AB* and carry it through to *K,* join *AK* and *KB,* draw *FL* from *F* perpendicular to *AK,* carry it through to *M,* and join *KN.*

Since the circumference *ABCG* equals the circumference *AEDG,* and in them *ABC* equals *AED,* therefore the remainder, the circumference *CG,* equals the remainder *GD.*

But *CD* belongs to a pentagon, therefore *CG* belongs to a decagon.

And, since *FA* equals *FB,* and *FH* is perpendicular, therefore the angle *AFK* equals the angle *KFB.*

Hence the circumference *AK* equals *KB.* Therefore the circumference *AB* is double the circumference *BK.* Therefore the straight line *AK* is a side of a decagon. For the same reason *AK* is double *KM.*

Now, since the circumference *AB* is double the circumference *BK,* while the circumference *CD* equals the circumference *AB,* therefore the circumference *CD* is also double the circumference *BK.*

But the circumference *CD* is also double *CG,* therefore the circumference *CG* equals the circumference *BK.* But *BK* is double *KM,* since *KA* is so also, therefore *CG* is also double *KM.*

But, further, the circumference *CB* is also double the circumference *BK,* for the circumference *CB* equals *BA.* Therefore the whole circumference *GB* is also double *BM.* Hence the angle *GFB* is double the angle *BFM.*

But the angle *GFB* is double the angle *FAB,* for the angle *FAB* equals the angle *ABF.* Therefore the angle *BFN* equals the angle *FAB.*

But the angle *ABF* is common to the two triangles *ABF* and *BFN,* therefore the remaining angle *AFB* equals the remaining angle *BNF.* Therefore the triangle *ABF* is equiangular with the triangle *BFN.*

Therefore, proportionally the straight line *AB* is to *BF* as *FB* is to *BN.* Therefore the rectangle *AB* by *BN* equals the square on *BF.*

Again, since *AL* equals *LK,* while *LN* is common and at right angles, therefore the base *KN* equals the base *AN.* Therefore the angle *LKN* also equals the angle *LAN.*

But the angle *LAN* equals the angle *KBN,* therefore the angle *LKN* also equals the angle *KBN.*

And the angle at *A* is common to the two triangles *AKB* and *AKN.* Therefore the remaining angle *AKB* equals the remaining angle *KNA.*

Therefore the triangle *KBA* is equiangular with the triangle *KNA.* Therefore, proportionally the straight line *BA* is to *AK* as *KA* is to *AN.*

Therefore the rectangle *BA* by *AN* equals the square on *AK.*

But the rectangle *AB* by *BN* was also proved equal to the square on *BF,* therefore the sum of the rectangle *AB* by *BN* and the rectangle *BA* by *AN,* that is, the square on *BA,* equals the sum of the squares on *BF* and *AK.*

And *BA* is a side of the pentagon, *BF* of the hexagon, and *AK* of the decagon.

Therefore, *if an equilateral pentagon is inscribed in a circle, then the square on the side of the pentagon equals the sum of the squares on the sides of the hexagon and the decagon inscribed in the same circle.*

Q.E.D.