Upon any side of the given polygon describe (E. 46. 1.) a square; find (S. 21. 6.) a square which shall have to the square first described the given ratio; and upon its side describe (E. 18. 6.) a polygon similar, and similarly situated, to the given polygon: It is manifest, from E. 20. 6., that it will have to the given polygon the given ratio.

PROP. LXVI.

75. THEOREM. Any regular polygon, inscribed in a circle, is a mean proportional between the inscribed and circumscribed regular polygons of half the number of sides.

cle BG, and let EH and CD be polygons of half the number of sides, the one EH inscribed in the

circle (S. 14. 4. cor. 3.) by joining the sides of the figure BGFA, and the other CD described about the circle, by drawing tangents to it through the angular points A, B, G, and F; so that (E. 18. 3. E. 28. 3. E. 27. 3. E. 26. 3. and S. 19. 3.) it is equilateral and equiangular: Then is the polygon BGFA a mean proportional between the polygons EH and CD.

Find (E. 1. 3.) the centre K of the circle BGFA, and join K, B, and K, C: It is manifest, from the construction, that KB bisects, at right, the sides of the figures EH and CD, which it cuts, and that KC passes through the angular point E: And (E. 1. 6. and E. 4. 6.),

A CBK: A EBK:: CK: EK::CB: EL: But, the figure CD is the same multiple of the A CBK, that the figure BGFA is of the ▲ EBK; also a side of CD is double of CB; and a side of EH is double of EL; .. (E. 15. 5.) CD is to BGFA as a side of CD is to a side of EH; and (E. 20. 6.) CD has to EH the duplicate ratio, of that which a side of CD has to a side of EH ;

CD has to EH the duplicate ratio, of that which it has to BGFA; i. e. (E. 10. def. 5.) the figure BGFA is a mean proportional between CD and EH.

PROP. LXVII.

76. THEOREM. If from two points similarly situated, one in each of any two homologous sides of two similar polygons, two straight lines be drawn making equal angles with those sides, they shall cut off from the polygons two similar figures; and the one shall be the same part of the one polygon, that the other is of the other.

Let AC and FH be two similar polygons, and P

and Q two points similarly situated in the two homologous sides CD and HK: If from P and Q straight lines be drawn, making equal with CD and HK, they shall cut off similar figures from the polygons; and the one shall be the same part of the one polygon that the other is of the other.

First, let PM and QN, making the MPD = 4 NQK, cut the sides DE and KL, adjacent to

CD and HK: And since (hyp. and S. 26. 1.) the two MPD, NQK, are equiangular, they are (E. 4. 6.) similar to one another, and they are to one another (E. 19. 6.) in the duplicate ratio of their homologous sides PD and QK, that is (hyp.) in the duplicate ratio of CD and HK; .. (E. 20. 6.) they are to one another in the same ratio as the polygons are, and .. whatever part the A MPD is of the polygon ABCDE, the same part is the ▲ NQK of the polygon FGHKL.

Secondly, let PR and QS cut any other sides of the polygons, as AE and FL, which are not adjacent to the sides CD and HK: Draw PE and QL; and since (hyp. and E. 26. 1.) the EPD and LQK are equiangular, the RPE and SQL are also equiangular; whence it may be shewn, (as in E. 20. 6.) that RPDE, SQKL, are similar figures; .. (E. 20. 6.) they are to one another in the duplicate ratio of the homologous sides DE and KL; or in the ratio of the polygon ABCDE to the polygon FGHKL; .. RPDE is the same part of ABCDE that SQKL is of FGHKL.

PROP. LXVIII.

77. THEOREM. If any two chords of a circle intersect each other, the straight lines joining their extremities shall cut off equal segments from the chord which passes through the common inter

section of the two former chords and is there bisected.

Let AB and CD be two chords of the circle

ACBD, cutting one another in E; through E draw (S. 2. 3.) the chord FG, so that FG is bisected in E; and join C, B and A, D; Then shall HE=EI.

For through I draw (E. 31. 1.) KIL parallel to BC, and meeting CD in K, and BA, produced, in L: Then (constr. and E. 29. 1.) the CBL: < BLK, and that (E. 21. 3.) the CBA = L CDA, the ALI = 2 IDK; and (E. 15. 1.) the AIL, of the ▲ LAI, is equal to

the

KID,

are

of the ▲ DKI; .. (S. 26. 1.) these two equiangular, as are also (constr. and E. 29. 1.) the two CEH, IEK, and the two IEL;

.. (E. 4. 6.) AI:IL::KI:ID;

.. (E. 16. 6.) IL × KI = AI × ID:

HEB,