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EXERCISES.

on PROPOSITION 1.

1. If two circles intersect at the points A, B, shew that the line which joins their centres bisects their common chord AB at right angles.

2. AB, AC are two equal chords of a circle ; shew that the straight line which bisects the angle BAC passes through the centre.

3. Two chords of a circle are given in position and magnitude : find the centre of the circle.

4. Describe a circle that shall pass through three given points, which are not in the same straight line.

5. Find the locus of the centres of circles which pass through two given points.

6. Describe a circle that shall pass through two given points, and have a given radius. When is this impossible ?

ON PROPOSITION 2.

7. A straight line cannot cut a circle in more than two points.

ON PROPOSITION 3.

8. Through a given point within a circle draw a chord which shall be bisected at that point.

9. The parts of a straight line intercepted between the circumferences of two concentric circles are equal.

10. The line joining the middle points of two parallel chords of a circle passes through the centre.

11. Find the locus of the middle points of a system of parallel chords drawn in a circle.

12. If two circles cut one another, any two parallel straight lines drawn through the points of intersection to cut the circles, are equal.

13. PQ and XY are two parallel chords in a circle: shew that the points of intersection of PX, QY, and of PY, QX, lie on the straight line which passes through the middle points of the given chords,

PROPOSITION 5. THEOREM.

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

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Let the two OS AGC, BFC cut one another at C.

Then they shall not have the same centre. Construction. If possible, let the two circles have the same centre; and let it be called E.

Join EC;

and from E draw any st. line to meet the O ces at F and G.

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Proof. Because E is the centre of the O AGC,

Нур. .: EG= EC. And because E is also the centre of the O BFC, Hyp.

EF= EC.

EG= EF,
the whole equal to its part, which is impossible.
Therefore the two circles have not the same centre.

..

Q.E.D.

EXERCISES.

ON PROPOSITIONS 4 AND 5. 1. If a parallelogram can be inscribed in a circle, the point of intersection of its diagonals must be at the centre of the circle.

2. Rectangles are the only parallelograms that can be inscribed in a circle.

3. Two circles, which intersect at one point, must also intersect at another.

PROPOSITION 6. THEOREM. If two circles touch one another internally, they cannot have the same centre.

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Let the two OS ABC, DEC touch one another internally at C.

Then they shall not have the same centre. Construction. If possible, let the two circles have the same centre; and let it be called F.

Join FC; and from F draw any st, line to meet the ces at E and B.

Нур.

Proof. Because F is the centre of the O ABC,

.. FB=FC.
And because F is the centre of the O DEC,

i'. FE=FC.

Нур.

FB=FE,
the whole equal to its part, which is impossible.
Therefore the two circles have not the same centre.

Q.E.D.

NOTE. From Propositions 5 and 6 it is seen that circles, whose circumferences have any point in common, cannot be concentric, unless they coincide entirely.

Conversely, the circumferences of concentric circles can have no point in common.

PROPOSITION 7. THEOREM.

If from any point within a circle. which is not the centre, straight lines are drawn to the circumference, then the greatest is that which passes through the centre; and the least is the remaining part of the diameter.

And of all other such lines, that which is nearer to the greatest is always greater than one more remote.

And two equal straight lines, and only two, can be drawn from the given point to the circumference, one on each side of the diameter.

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Let ABCD be a circle, and from F, any point within it which is not the centre, let FA, FB, FC, FG, and FD be drawn to the oce, of which FA passes through E the centre, and FD is the remaining part of the diameter.

Then of all these st. lines, (i) FA shall be the greatest ; (ii) FD shall be the least ; (iii) FB, which is nearer to FA, shall be greater than FC,

which is more remote ; (iv) also two, and only two, equal st. lines can be drawn from

F to the oce.
Construction. Join EB, EC.

Proof. (i) In the A FEB, the two sides FE, EB are together greater than the third side FB.

I. 20. But EB=EA, being radii of the circle ; .. FE, EA are together greater than FB; that is, FA

greater than FB.

се

Similarly FA may be shewn to be greater than any other st. line drawn from F to the oce;

.. FA is the greatest of all such lines. (ii) In the A EFG, the two sides EF, FG are together greater than EG;

I. 20. and EG=ED, being radii of the circle ; .: EF, FG are together greater than ED. Take away the common part EF;

then FG is greater than FD. Similarly any other st. line drawn from F to the oce may be shewn to be greater than FD;

.:: FD is the least of all such lines.
(iii)

In the AS BEF, CEF,
BENCE,

1. Def. 15. Because

and EF is common ;
but the 2 BEF is greater than the 2 CEF;
i. FB is greater than FC.

I. 24. Similarly it may be shewn that FC is greater than FG.

(iv) Join EG, and at E in FE make the 2 FEH equal to the L FEG.

I. 23.
Join FH.
Then in the AS GEF, HEF,
GE=HE,

1. Def. 15.
and EF is common ;
(also the 2 GEF=the 2 HEF; Constr.
.. FG= FH.

I. 4. And besides FH no other straight line can be drawn from F to the O ce equal to FG.

For, if possible, let FK = FG.
Then, because FH = FG,

Proved. ... FK= FH, that is, a line nearer to FA, the greatest, is equal to a line which is more remote ; which is impossible. Proved.

Therefore two, and only two, equal st. lines can be drawn from F to the oce.

Q.E.D.

Because

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