PROPOSITION VII.

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

Let APB, AQB be angles in the same segment APQB. Then shall the LS APB, AQB be equal to one another.

First, let the segment APQB be greater than a semicircle.

Join A and B with C the centre of the circle of which APB is a segment.

Then ACB is double of each of the LS APB,AQB;

.. Ls APB, AQB are equal to one another. Next, let the segment APQB

be not greater than a semicircle.

Draw PCH through the centre C, and join HQ.

Then the S APH, AQH are in a segment greater than a semicircle, and are .. equal to one another.

H

(III. 5)

B

Similarly s BPH, BQH are equal to one another.

... the whole APB is Wherefore the angles &c.

=

the whole AQB.

Q.E.D.

PROPOSITION VIII.

In equal circles equal angles stand upon equal

arcs.

M

Let AFM, BHN be equal circles, and let the s FPG, HQK at their centres be equal to one another. Then shall the arc FMG be equal to the arc HNK.

For let the circle AFM be applied to the circle BHN, so that P may fall on Q,

and the circumference AFM on the circumference BHN;

then if PF fall on QH,

=

PG will fall on QK, since ▲ FPG is .. the arc FMG will fall on the arc HNK, and is therefore equal to it.

Wherefore in equal circles &c.

L HQK.

Q.E.D.

COROLLARY.—In equal circles, or in the same circle,

equal chords cut off equal arcs.

For if the chord FG be

=

the chord HK ;

since also PF and PG are respectively equal to QH

ON THE SUBDIVISION

OF THE CIRCUMFERENCE OF A CIRCLE.

PROPOSITION IX.

To bisect a given arc of a circle.

H

Let AXB be the given arc of a circle.

It is required to bisect it.

Join AB and bisect it in H.

Through H draw HX perpendicular to AB,

cutting the arc in X.

The arc AXB shall be bisected in X.

Join AX and XB.

(I. 9)

(I. 12)

Because AH = HB and HX is common to the As

PROPOSITION X.

To bisect the circumference of a circle.

Let AFB be the given circle.
It is required to bisect its circumference.

Through the centre C draw any diameter ACB, then shall the circumference be bisected by ACB.

For if the circle were taken up, inverted, and then applied to its former position so that ACB should fall on its former position,

then would the arc AGB fall upon the arc AFB;

.. arc AGB

=

arc AFB.

Wherefore the circumference &c.

Q.E. F.

COROLLARY.-Hence the circumference of a circle may be divided into 4, 8, 16, 32, &c., equal arcs.

For if through C, FCG be drawn perpendicular to ACB, then FCG will bisect the arcs AGB, AFB; (III. 9) .. the circumference will be divided into four equal parts. Again, if the angles at the centre be bisected,

then each of the arcs AG, GB, BF, FA will be bisected, (III. 8)

and .. the circumference will be divided into eight equal parts.

And so on.