101. Fig. 39 shows how the pentedecagon is obtained from the pentagon.

Let ABCDE be the pentagon and O its center.

Fig. 39.

Draw OA, OB, OC, OD, and OE. Produce DO to meet AB in K.

Take OF=½ of OD.

Fold GFH at right angles to OF. Make OG = OH= OD.

Then GDH is an equilateral triangle, and the angles DOG and HOD are each 120°.

But angle DOA is 144°; therefore angle GOA is 24°.

That is, the angle EOA, which is 72°, is trisected by OG.

Bisect the angle EOG by OL, meeting EA in L, and let OG cut EA in M; then

OL = OM.

In OA and OE take OP and OQ equal to OL or OM.

Then PW, ML, and LQ are three sides of the pentedecagon.

Treating similarly the angles A OB, BOC, COD, and DOE, we obtain the remaining sides of the pentedecagon.