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PROPOSITION XVII. PROBLEM.
394. To inscribe in a given circle a regular decagon.
Let O be the centre of the given circle.
Draw the radius 0 C, and divide it in extreme and mean ratio, so that 0 C shall be to O S as 0 S is to SC.
Draw BC, BS, and B 0.
Cons. Substitute for OS its equal B C, then OC : BC :: BC:SC.
Moreover the LOCB=ZSC B, Iden.
.. the A OC B and B C S are similar, $ 284 (having an Z of the one equal to an Zof the other, and the including sides
§ 160 (its sides 0 C and 0 B being radii of the same circle). .. the A BCS, which is similar to the AOC B, is isosceles,
Cons. .:.08= BS,
Ax. 1 .. the A S O B is isosceles, and the 20=2 SBO,
§ 112 (being opposite equal sides).
But the ZC SB=20+ Z SBO, § 105 (the exterior 2 of a A is equal to the sum of the two opposite interior &).
... the Z CSB = 2 2 0.
ZSCB=2 C SB) = 2 2 0, $ 112 and ZOBC=2 SC B) = 2 20. § 112 is the sum of the 4s of the A OCB=5 20. .:15 20= 2 rt. 6,
§ 98 and 20= 1 of 2 rt. £, or 1 of 4 rt. E.
.. the arc B C is to of the circumference, and
.. the chord B C is a side of a regular inscribed decagon.
Hence, to inscribe a regular decagon, divide the radius in extreme and mean ratio, and apply the greater segment ten times as a chord.
Q. E. F.
395. COROLLARY 1. By joining the alternate vertices of a regular inscribed decagon, a regular pentagon may be inscribed.
396. Cor. 2. By bisecting the arcs BC, CF, etc., a regular polygon of 20 sides may be inscribed, and, by continuing the process, regular polygons of 40, 80, etc., sides may be inscribed. PROPOSITION XVIII. PROBLEM. 397. To inscribe in a given circle a regular pentedecagon, or polygon of fifteen sides.
Let Q be the given circle. It is required to inscribe in Q a regular pentedecagon. Draw E H equal to a side of a regular inscribed hexagon, § 391 and E F equal to a side of a regular inscribed decagon. $ 394
and the arc E F is Ib of the circumference;
.. the chord FH is a side of a regular inscribed pentedecagon,
and by applying FH fifteen times as a chord, we have the polygon required.
Q.E.F. 398. COROLLARY. By bisecting the arcs FH, HA, etc., a regular polygon of 30 sides may be inscribed ; and by continuing the process, regular polygons of 60, 120, etc. sides may be inscribed.
PROPOSITION XIX. PROBLEM. 399. To inscribe in a given circle a regular polygon similar to a given regular polygon. CDI
Let A B C D, etc., be the given regular polygon, and
C'D' E' the given circle.
It is required to inscribe in C'D' E' a regular polygon. similar to A B C D, etc. From O, the centre of the polygon A B C D, etc.
draw 0 D and 0 C.
draw O'C' and O'D', making the 20' = < 0.
Draw C' D'. Then C' D' will be a side of the regular polygon required.
For each polygon will have as many sides as the 20 (=20') is contained times in 4 rt. 4.
.. the polygon C' D' E', etc. is similar to the polygon C D E, etc.,
§ 372 (two regular polygons of the same number of sides are similar).
Q. E. F.
PROPOSITION XX. PROBLEM.
400. To circumscribe about a circle a regular polygon similar to a given inscribed regular polygon.
Let H MRS, etc., be a given inscribed regular polygon.
It is required to circumscribe a regular polygon similar to HM RS, etc.
At the vertices H, M, R, etc., draw tangents to the O, intersecting each other at A, B, C, etc.
Then the polygon A B C D, etc. will be the regular polygon required.
Since the polygon A B C D, etc.
has the same number of sides as the polygon H MRS, etc.,
it is only necessary to prove that A B C D, etc. is a regular polygon.
In the A BHM and CMR,