To circumscribe an equilateral and equiangular pentagon about a given circle.

Let *ABCDE* be the given circle.

It is required to circumscribe an equilateral and equiangular pentagon about the circle *ABCDE.*

Let *A, B, C, D,* and *E* be conceived to be the angular points of the inscribed pentagon, so that the circumferences *AB, BC, CD,* *DE,* and *EA* are equal.

Draw *GH, HK, KL, LM,* and *MG* through *A, B, C, D,* and *E* touching the circle. Take the center *F* of the circle *ABCDE,* and join *FB, FK, FC, FL,* and *FD.*

Then, since the straight line *KL* touches the circle *ABCDE* at *C,* and *FC* has been joined from the center *F* to the point of contact at *C,* therefore *FC* is perpendicular to *KL.* Therefore each of the angles at *C* is right.

For the same reason the angles at the points *B* and *D* are also right.

And, since the angle *FCK* is right, therefore the square on *FK* equals the sum of the squares on *FC* and *CK.*

For the same reason the square on *FK* also equals the sum of the squares on *FB* and *BK,* so that the sum of the squares on *FC* and *CK* equals the sum of the squares on *FB* and *BK,* of which the square on *FC* equals the square on *FB,* therefore the remaining square on *CK* equals the square on *BK.*

Therefore *BK* equals *CK.*

And, since *FB* equals *FC,* and *FK* is common, the two sides *BF* and *FK* equal the two sides *CF* and *FK,* and the base *BK* equals the base *CK,* therefore the angle *BFK* equals the angle *KFC,* and the angle *BKF* equals the angle *FKC.* Therefore the angle *BFC* is double the angle *KFC,* and the angle *BKC* double the angle *FKC.*

For the same reason the angle *CFD* is also double the angle *CFL,* and the angle *DLC* double the angle *FLC.*

Now, since the circumference *BC* equals *CD,* the angle *BFC* also equals the angle *CFD.*

And the angle *BFC* is double the angle *KFC,* and the angle *DFC* double the angle *LFC,* therefore the angle *KFC* also equals the angle *LFC.*

But the angle *FCK* also equals the angle *FCL,* therefore *FKC* and *FLC* are two triangles having two angles equal to two angles and one side equal to one side, namely *FC* which is common to them, therefore they will also have the remaining sides equal to the remaining sides, and the remaining angle to the remaining angle, therefore the straight line *KC* equals *CL,* and the angle *FKC* equals the angle *FLC.*

And, since *KC* equals *CL,* therefore *KL* is double *KC.*

For the same reason it can be proved that *HK* is also double *BK.*

And *BK* equals *KC,* therefore *HK* also equals *KL.*

Similarly each of the straight lines *HG, GM,* and *ML* can also be proved equal to each of the straight lines *HK* and *KL,* therefore the pentagon *GHKLM* is equilateral.

I say next that it is also equiangular.

For, since the angle *FKC* equals the angle *FLC,* and the angle *HKL* was proved double the angle *FKC,* and the angle *KLM* double the angle *FLC,* therefore the angle *HKL* also equals the angle *KLM.*

Similarly each of the angles *KHG, HGM,* and *GML* can also be proved equal to each of the angles *HKL* and *KLM.* Therefore the five angles *GHK, HKL,* *KLM, LMG,* and *MGH* equal one another.

Therefore the pentagon *GHKLM* is equiangular.

And it was also proved equilateral, and it has been circumscribed about the circle *ABCDE.*

Q.E.F.

Conversely, if you have a regular circumscribed *n*-gon, then you can connect the points of tangency in sequence to get a regular inscribed *n*-gon.