PROPOSITION II.

The straight line drawn through the middle point of a chord of a circle and the centre is perpendicular to the chord.

Let AB be a chord of a of which C is the centre. Then the straight line CH through C and the middle point H of AB shall be to AB.

Join CA, CB.

Then AH, HC, CA are respectively BH, HC, CB;

...

LAHC= L BHC;

.. CH is 1 AB.

(1.5)

Conversely, The straight line drawn through the centre of a circle to a chord bisects the chord.

For through the centre only one straight line can be drawn to the chord; and that through the middle point of the chord is to it.

Also, The straight line drawn through the middle point of a chord of a circle to the chord passes through the centre. For through the middle point of the chord only one straight line can be drawn to it; and that through the centre is to it.

COR. The middle point of a chord bisecting another at right angles is the centre.

PROPOSITION III.

If from a point within a circle there be drawn to the circumference more than two equal straight lines, that point is the centre of the circle.

from which are drawn to the circumference the equal straight lines SP, SQ, SR.

S shall be the centre of the circle.

Join PQ, QR; bisect PQ in H and QR in K.

Draw a straight line through H, S,

also a straight line through K, S.

Then. PH, HS, SP are respectively QH, HS, SQ;

.. the centre of the is in the straight line HS. (III. 2)

Similarly it is in the straight line KS;

.. S is the centre of the O PQR.

C. G.

7

DEFINITION.

Straight lines are said to be equally distant from a given point when the perpendiculars let fall upon them from the given point are equal; also one straight line is said to be farther than another from a given point when the perpendicular from the given point on the former is greater than that on the latter.

PROPOSITION IV.

Equal chords of a circle are equally distant from the centre.

B

Ꭱ

Let AB, CD be equal chords of the

ABC.

Then shall AB, CD be equally distant from the centre Q.

Draw QR to AB and.. bisecting AB, (III. 2) and QSL to CD and ... bisecting CD. (111. 2) Join AQ, CQ.

Then ·.· AB = CD; .. AR = CS;

also. AQ CQ; .. square on AQ= square on CQ;

=

... squares on AR, RQ = squares on CS, SQ, (1. 35)

but square on AR = square on CS;

.. square on RQ = square on SQ;`
.. RQ = SQ;

.. AB, CD are equally distant from the centre of ABC.

Conversely:

Chords in a circle which are equally distant from the centre are equal to one another.

Let the chords AB, CD of the ABC be equally distant from the centre Q;

that is, let the perpendiculars QR, QS drawn to them from Q be equal.

Then shall AB, CD be equal to one another.

Join Q4, QC.

Then . QR, QS are to AB, CD respectively;
.. AB, CD are bisected in R, S.
QA = QC, ... square on QA = square on QC;

Also

(III. 2)

... squares on QR, RA are together = squares on QS, SC, (1.35)

but square on QR = square on QS; '.' QR = QS ;
... square on RA = square on SC;

... RA= SC;

.. AB = CD.

DEFINITION.

A chord of a circle which passes through the centre is called a diameter.

PROPOSITION V.

The diameter is the greatest chord of a circle, and of all others that nearer to the centre is greater than the more remote.

HK, FG chords of which HK is nearer to the centre C

than FG.

Then shall AB be > HK and HK > FG.

Again,

Draw CX 1 to HK, and... bisecting HK. (111. 2)
Also CYL to FG, and.. bisecting FG.

Join CH, CK, CF.

Then HC, CK are together > HK ;

.. AC, CB are together > HK;

CH

=

i. e. ACB is > HK.

(1. 16)

CF; ... square on CH = square on CF; ... squares on CX, XH are together = squares on CY, YF; (1.35) but CX is < CY, and .. square on CX is < square on CY;

.. square on XH is > square on YF,

and ... XH is > YF;

.. HK is > FG.