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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.
It is required to inscribe in the given O a regular decagon.

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.

§ 311
From C as a centre, with a radius equal to OS,
describe an arc intersecting the circumference at B.

Draw BC, BS, and B 0.
Then B C is a side of the regular decagon required.
For
00:0S ::08 : SC,

Cons.
BC= 0 S.

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

proportional).
But the AOC B is isosceles,

§ 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,

and

and
BS=BC.

$ 114
But
OS=BC,

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

Join FH.
Then FH will be a side of a regular inscribed pentedecagon.
For the arc E H is of the circumference,

and the arc E F is Ib of the circumference;
.. the arc F H is - to, or 1, 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

D

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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.
From O' the centre of the OC" D' E,

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.

BMC

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,

§ 372

§ 363

HM= MR,
(being sides of a regular polygon),

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