Book 4
11
For, since the circumference AB is equal to the circumference DE, let BCD be added to each; therefore the whole circumference ABCD is equal to the whole circumference EDCB.
11
And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is also equal to the angle AED. [III. 27]
11
For the same reason each of the angles ABC, BCD, CDE is also equal to each of the angles BAE, AED; therefore the pentagon ABCDE is equiangular.
11
But it was also proved equilateral; therefore in the given circle an equilateral and equiangular pentagon has been inscribed. Q. E. F.
PROPOSITION 12.
12
About a given circle to circumscribe an equilateral and equiangular pentagon.
12
Let ABCDE be the given circle; thus it is required to circumscribe an equilateral and equiangular pentagon about the circle ABCDE.
12
Let A, B, C, D, E be conceived to be the angular points of the inscribed pentagon, so that the circumferences AB, BC, CD, DE, EA are equal; [IV. 11] through A, B, C, D, E let GH, HK, KL, LM, MG be drawn touching the circle; [III. 16, Por.] let the centre F of the circle ABCDE be taken [III. 1], and let FB, FK, FC, FL, FD be joined.
12
Then, since the straight line KL touches the circle ABCDE at C, and FC has been joined from the centre F to the point of contact at C, therefore FC is perpendicular to KL; [III. 18] therefore each of the angles at C is right.
12
For the same reason the angles at the points B, D are also right.
12
And, since the angle FCK is right, therefore the square on FK is equal to the squares on FC, CK.
12
For the same reason [I. 47] the square on FK is also equal to the squares on FB, BK; so that the squares on FC, CK are equal to the squares on FB, BK, of which the square on FC is equal to the square on FB; therefore the square on CK which remains is equal to the square on BK.
12
Therefore BK is equal to CK.
