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PROPOSITION XIV. PROBLEM.

To describe a circle about a given regular pentagon.

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Let ABCDE be the given regular pentagon.

It is required to describe a C about the pentagon.

Bisect the 4s BCD, CDE by the st. lines CO, DO, meeting in 0.

Join OB, OA, OE.

Then it may be shewn, as in the preceding Proposition, that OB, OA, OE bisect the 4s CBA, BAE, AED.

And BCD= ▲ CDE,

and OCD half ▲ BCD, and

ODC=half CDE,

:. ▲ OCD= ▲ ODC,

and .. OD=OC.

In the same way we may shew that OB, OA, OE

and a

each-OD or OC;

.. OA, OB, OC, OD, OE are all equal,

described with centre O and radius OA will pass

through B, C, D, E,

and will be described about the pentagon.

Q. E. F.

PROPOSITION XXIV. THEOREM.

Similar segments of circles, upon equal straight lines, are equal to one another.

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Let ABC, DEF be similar segments of Os on equal st. lines AB, DE.

Then must segment ABC=segment DEF.

For if segment ABC be applied to segment DEF, so that A may be on C and AB on DE, then B will coincide with E, and AB with DE;

.. segment ABC must also coincide with segment DEF;

.. segment ABC=segment DEF.

III. 23.

Ax. 8.

Q. E. D.

We gave one Proposition, C, page 150, as an example of the way in which the conceptions of Flat and Reflex Angles may be employed to extend and simplify Euclid's proofs. We here give the proofs, based on the same conceptions, of the important propositions xxII. and xxxI.

PROPOSITION XXII. THEOREM.

The opposite angles of any quadrilateral figure, inscribed in a circle, are together equal to two right angles.

D

Let ABCD be a quadrilateral fig. inscribed in a ©.

Then must each pair of its opposites be together equal to two rt. 4 s.

From O, the centre, draw OB, OD.

Then :: ▲ BOD=twice ▲ BAD,

and the reflex 4 DOB=twice ▲ BCD,

III. 20.

III. C. p. 150.

I. 15, Cor. 2.

.. sum of 4s at 0=twice sum of 4s BAD, BCD.

L

But sum of 4 s at 0=4 right s;

.. twice sum of 4 8 BAD, BCD=4 rights;

.. sum of 4s BAD, BCD=two rights.

Similarly, it may be shewn that

sum of 4s ABC, ADC=two rights.

Q. E. D.

PROPOSITION XXXI. THEOREM.

In a circle, the angle in a semicircle is a right angle; and the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

B

Let ABC be a O, of which O is the centre and BC a diameter.

Draw AC, dividing the

into the segments ABC, ADC.

Join BA, AD, DC.

and

Then must the 4 in the semicircle BAC be a rt. 4, and ▲ in segment ABC, greater than a semicircle, less than a rt. L in segment ADC, less than a semicircle, greater than a rt. ▲ .

First, the flat angle BOC=twice BAC,

III. C. p. 150.

.. BAC is a rt. .

Next, BAC is a rt. 4,

.. ABC is less than a rt. 4.

Lastly, sum of 4 s ABC, ADC=two rt. 4s,

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I. 17.

III. 22.

Q. E. D.

BOOK I V.

INTRODUCTORY REMARKS.

EUCLID gives in this Book of the Elements a series of Problems relating to cases in which circles may be described in or about triangles, squares, and regular polygons, and of the last-mentioned he treats of three only:

the Pentagon, or figure of 5 sides,

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The Student will find it useful to remember the following Theorems, which are established and applied in the proofs of the Propositions in this Book.

I. The bisectors of the angles of a triangle, square, or regular polygon meet in a point, which is the centre of the inscribed circle.

II. The perpendiculars drawn from the middle points of the sides of a triangle, square, or regular polygon meet in a point, which is the centre of the circumscribed circle.

III. In the case of a square, or regular polygon the inscribed and circumscribed circles have a common centre.

IV. If the circumference of a circle be divided into any number of equal parts, the chords joining each pair of consecutive points form a regular figure inscribed in the circle, and the tangents drawn through the points form a regular figure described about the circle.

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