We here insert Euclid's proofs of Props. 23, 24 of Book III. first observing that he gives the following definition of similar segments:

Def. Similar segments of circles are those in which the angles are equal, or which contain equal angles.

BOOK III. PROPOSITION XXIII. THEOREM.

Upon the same straight line, and upon the same side of it, there cannot be two similar segments of circles, not coinciding with each other.

D

If it be possible, on the same base AB, and on the same side of it, let there be two similar segments of Os, ABC, ABD, which do not coincide.

Because ADB cuts ACB in pts. A and B, they cannot cut one another in any other pt., and .. one of the segments must fall within the other.

Let ADB fall within ACB.

Draw the st. line BDC and join CA, DA.

Then. segment ADB is similar to segment ACB,

.. LADB= LACB.

Or the extr. 4 of a ▲ the intr. and opposite 4,

impossible;

=

which is

.: the segments cannot but coincide.

Q. E. D.

BOOK III. PROPOSITION XXIV. THEOREM.

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

Let ABC, DEF be similar segment 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 opposite s be together equal to two rt. LS.

From O, the centre, draw OB, OD.

Then

BOD=twice ▲ BAD,

III. 20.

and the reflex 4 DOB=twice ▲ BCD, III. C. p. 150.

.. sum of 8 at 0=twice sum of 48 BAD, BCD.

But sum of 48 at 0=4 right s;

1. 15, Cor. 2.

.. twice sum of 48 BAD, BCD=4 right 48;

.. sum of 48 BAD, BCD=two right 4 s.

Similarly, it may be shewn that

sum of 48 ABC, ADC=two right ▲ 8.

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 les than a right angle; and the angle in a segment less than semicircle is greater than a right angle.

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

Draw AC, dividing the into the segments ABC, ADC.
Join BA, AD, DC.

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

First, the flat angle BOC=twice BAC, III. C. p. 150. .. BAC is a rt. 4.

Next,: BAC is a rt. 4,

. LABC is less than a rt. 4.

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

I. 17.

III. 22.

and ABC is less than a rt. 4,

.. LADC is greater than a rt. 2.

Q. E. D.