Next, suppose to be within (fig 1), or without (fig. 2) the L BAC.

Join 40, and produce it to meet the Oce in D. Then, as in the first case,

▲ COD = twice ▲ CAD,

and ▲ BOD = twice ▲ BAD ;

.., fig. 1, sum of 4 s COD, BOD = twice sum of 48 CAD, BAD,

that is, ▲ BOC = twice ▲ BAC.

And, fig. 2, difference of 4 s COD, BOD = twice difference of 48 CAD, BAD, that is, ▲ BOC = twice ▲ BAC.

Q. E. D.

Ex. 1. The centre of the circle CBED is on the circumference of ABD. If from any point A the lines ABC and AED be drawn to cut the circles, the chord BE is parallel to CD.

Ex. 2. From any point in a straight line, touching a circle, a straight line is drawn through the centre, and is terminated by the circumference; the angle between these two straight lines is bisected by a straight line, which intersects the straight line joining their extremities. Shew that the angle between the last two lines is half a right angle.

NOTE 2. On Flat and Reflex Angles.

We have already explained (Note 3, Book I., p. 28) how Euclid's definition of an angle may be extended with advantage, so as to include the conception of an angle equal to two right angles and we now proceed to shew how the Definition given in that Note may be extended, so as to embrace angles greater than two right angles.

Let WQ be a straight line, and QE its continuation.

Then, by the Definition, the angle made by WQ and QE, which we propose to call a FLAT ANGLE, is equal to two right angles.

Now suppose QP to be a straight line, which revolves about the fixed point Q, and which at first coincides with QE.

When QP, revolving from right to left, coincides with QW, it has described an angle equal to two right angles.

When QP has continued its revolution, so as to come into the position indicated in the diagram, it has described an angle EQP, indicated by the dotted line, greater than two right angles, and this we call a REFLEX ANGLE.

To assist the learner, we shall mark these angles with dotted lines in the diagrams.

Admitting the existence of angles, equal to and greater than two right angles, the Proposition last proved may be extended, as we now proceed to shew.

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.

Fig. 2.

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.

Join DO, and produce it to meet the arc ACB in C.

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

Q. E. D.

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.

PROPOSITION XXI. THEOREM.

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

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.

Next, when segment BADC is less than a semicircle,

(Fig. 1.)

III, 20.

III. 20.

Let E be the pt. of intersection of AC, DB. (Fig. 2.)
ABE= ▲ DCE, by the first case,

Then.

and ▲ BEA= 4 CED,

L

.. LEAB= LEDC,

that is, BAC= ▲ BDC.

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

Let ABCD be a quadrilateral fig. inscribed in a .

Then must each pair of its opposite s be together equal to

two rt. 4 s.

Draw the diagonals AC, BD.

Then : ADB= ▲ ACB, in the same segment,

III. 21.

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

III. 21.

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

that is, ≤ ADC=sum of ▲ 8 ACB, BAC.

Add to each ▲ ABC.

L

Then 48 ADC, ABC together sum of 4s ACB, BAC, ABC;

and .. 4 8 ADC, ABC together=two right ▲ s. Similarly, it may be shewn,

I. 32.

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

Q. E. D.

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