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PROPOSITION C. THEOREM.

The angle, not less than two right angles, at the centre of a circle is double of the angle at the circumference, subtended by the same arc.

Fig. 1.

B

Fig. 2.

A

In the ACBD, let the angles AOB (not less than two right angles) at the centre, and ADB at the circumference, be subtended by the same arc ACB.

Then must AOB=twice ▲ ADB.

L

Join DO, and produce it to meet the arc ACB in C.
AOC=twice ▲ ADO,

Then

III. 20.

and 4 BOC=twice / BDO,

III. 20.

.. sum of 4 s AOC, BOC=twice sum of 2 8 ADO, BDO, that is, ▲ AOB=twice ▲ ADB.

Q. E. D.

NOTE. In fig. 1, 2 AOB is drawn a flat angle,

and in fig. 2, 4 AOB is drawn a reflex angle.

DEF. XII. The angle in a segment is the angle contained by two straight lines drawn from any point in the arc to the extremities of the chord.

Fig. 1.

PROPOSITION XXI. THEOREM.

The angles in the same segment of a circle are equal to one another.

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Let BAC, BDC be angles in the same segment BADC.
Then must BAC= L BDC.

First, when segment BADC is greater than a semicircle,
From 0, the centre, draw OB, OC.
Then, . BOC=twice / BAC,

and BOC=twice 4 BDC,
..L BAC=▲ BDC.

Next, when segment BADC is less than a semicircle,
Let E be the pt. of intersection of AC, DB. (Fig. 2.)
Then . ABE= ▲ DCE, by the first case,
and ▲ BEA= ▲ CED,

.. LEAB=L EDC,

that is, BAC=▲ BDC.

(Fig. 1.) III, 20.

III. 20.

I. 15.

I. 32.

Q. E. D.

Ex. 1. Shew that, by assuming the possibility of an angle being greater than two right angles, both the cases of this proposition may be included in one.

Ex. 2. AB, AC are chords of a circle, D, E the middle points of their arcs. If DE be joined, shew that it will cut off equal parts from AB, AC.

Ex. 3. If two straight lines, whose extremities are in the circumference of a circle, cut one another, the triangles formed by joining their extremities are equiangular to each other.

PROPOSITION XXII. THEOREM.

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

B

Draw the diagonals AC, BD.

Then

Let ABCD be a quadrilateral fig. inscribed in a O.
Then must each pair of its opposite s be together equal to

two rt. 4 s.

ADB

ACB, in the same segment,

and ▲ BDC= ▲ BAC, in the same segment;

.. sum of 4 s ADB, BDC=sum of 2 s ACB, BAC;

that is, ▲ ADC=sum of ▲ 8 ACB, BAC.

Add to each ▲ ABC.

III. 21.

III. 21.

and.. 48 ADC, ABC together=two right ▲ s. Similarly, it may be shewn,

that s BAD, BCD together=two right 4 s.

Then 48 ADC, ABC together=sum of 48 ACB, BAC, ABC;

I. 32,

Q. E. D.

NOTE. Another method of proving this proposition is given. on page 177.

Ex. 1. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the opposite angle of the quadrilateral.

Ex. 2. If the sides AB, DC of a quadrilateral inscribed in a circle be produced to meet in E, then the triangles EBC, EAD will be equiangular.

Ex. 3. Shew that a circle cannot be described about a rhombus.

Ex. 4. The lines, bisecting any angle of a quadrilateral figure inscribed in a circle and the opposite exterior angle, meet in the circumference of the circle.

Ex. 5. AB, a chord of a circle, is the base of an isosceles triangle, whose vertex C is without the circle, and whose equal sides meet the circle in D, E: shew that CD is equal to CE.

Ex. 6. If in any quadrilateral the opposite angles be together equal to two right angles, a circle may be described about that quadrilateral.

Propositions XXIII. and XXIV., not being required in the method adopted for proving the subsequent Propositions in this book, are removed to the Appendix. Proposition xxv. has been already proved.

NOTE 3. On the Method of Superposition, as applied
to Circles.

In Props. XXVI. XXVII. XXVIII. XXIX. we prove certain relations existing between chords, arcs, and angles in equal circles. As we shall employ the Method of Superposition, we must state the principles which render this method applicable, as a test of equality, in the case of figures with circular boundaries.

B

DEF. XIII. Equal circles are those, of which the radii are equal.

B

C

For suppose ABC, A'B'C' to be circles, of which the radii are equal.

Then if © A'B'C' be applied to © ABC, so that ơ, the centre of A'B'C', coincides with O, the centre of ABC, it is evident that any particular point A' in the Oce of the former must coincide with some point A in Oce of the latter, because of the equality of the radii O'A' and OA.

Hence Oce A'B'C' must coincide with Oce ABC,
that is, A'B'C' = © ABC.

Further, when we have applied the circle A'B'C' to the circle ABC, so that the centres coincide, we may imagine ABC to remain fixed, while A'B'C' revolves round the common centre. Hence we may suppose any particular point B' in the circumference of A'B'C' to be made to coincide with any particular point B in the circumference of ABC.

Again, any radius O'A' of the circle A'B'C' may be made to coincide with any radius OA of the circle ABC.

Also, if A'B' and AB be equal arcs, they may be made to coincide.

Again, every diameter of a circle divides the circle into equal segments.

For let AOB be a diameter of the circle ACBD, of which O is the centre. Suppose the segment ACB to be applied to the segment ADB, so as to keep AB a common boundary: then the arc ACB must coincide with the arc ADB, because every point in each is equally distant from 0.

B

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