PROPOSITION III. THEOREM.

If a straight line, drawn through the centre of a circle, bisect a chord of the circle, which does not pass through the centre, it must cut it at right angles: and conversely, if it cut it at right angles, it must bisect it.

In the

ABC, let the chord AB, which does not pass

through the centre O, be bisected in E by the diameter CD.

Then must CD be 1 to AB.

Join OA, OB.

Then in As AEO, BEO,

·· AE=BE, and EO is common, and OA=0B,

Next let CD be 1 to AB.

Then must CD bisect AB.

For OA=0B, and OE is common,

in the right-angled As AEO, BEO,

.. AE=BE,

that is, CD bisects AB.

I. E. Cor. p. 43.

Q. E. D.

Ex. 1. Shew that, if CD does not cut AB at right angles, it cannot bisect it.

Ex. 2. A line, which bisects two parallel chords in a circle, is also perpendicular to them.

Ex. 3. Through a given point within a circle, which is not the centre, draw a chord which shall be bisected in that point.

PROPOSITION IV. THEOREM.

If in a circle two chords, which do not both pass through the centre, cut one another, they do not bisect each other.

E

Let the chords AB, CD, which do not both pass through the centre, cut one another, in the pt. E, in the ACBD.

Then AB, CD do not bisect each other.

If one of them pass through the centre, it is plainly not bisected by the other, which does not pass through the centre.

But if neither pass through the centre, let, if it be possible, AE=EB and CE=ED; find the centre O, and join OE.

Then OE, passing through the centre, bisects AB,

.. LOEA is a rt. 4.

And OE, passing through the centre, bisects CD,

LOEA

= ▲ OEC, which is impossible;

.. AB, CD do not bisect each other.

III. 3.

III. 3.

Q. E. D.

Ex. 1. Shew that the locus of the points of bisection of all parallel chords of a circle is a straight line.

Ex. 2. Shew that no parallelogram, except those which are rectangular, can be inscribed in a circle.

PROPOSITION V. THEOREM.

If two circles cut one another, they cannot have the same centre.

If it be possible, let O be the common centre of the ABC, ADC, which cut one another in the pts. A and C. Join OA, and draw OEF meeting the Os in E and F. Then is the centre of O ABC,

.. OE=OA ;

and. O is the centre of o ADC,

.. OF=OA;

.. OE=OF, which is impossible ;
.. O is not the common centre.

I. Def. 13,

I. Def. 13.

Q. E. D.

Ex. Two circles, whose centres are A and B, intersect in C; through C two chords DCE, FCG are drawn equally inclined to AB and terminated by the circles: prove that DE and FG are equal.

NOTE. Circles which have the same centre are called Concentric.

NOTE 1. On the Contact of Circles.

DEF. VII. Circles are said to touch each other, which meet but do not cut each other.

One circle is said to touch another internally, when one point of the circumference of the former lies on, and no point without, the circumference of the other.

Hence for internal contact one circle must be smaller than the other.

Two circles are said to touch externally, when one point of the circumference of the one lies on, and no point within the circumference of the other.

N.B. No restriction is placed by these definitions on the number of points of contact, and it is not till we reach Prop. XIII. that we prove that there can be but one point of contact.

PROPOSITION VI. THEOREM.

If one circle touch another internally, they cannot have the same centre.

and let A be a point of contact.

Then some point E in the Oce ADE lies within ABC.

Def. 7.

If it be possible, let O be the common centre of the two Os. Join OA, and draw OEC, meeting the Oces in E and C.

.. O is not the common centre of the two Os.

Q. E. D.