PROPOSITION 9.

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

Let there be a pt. D within the
ABC such that three equal

st. lines DA, DB, DC can be
drawn from it to the Oce; then

D shall be the centre.

B

Join AB, BC; bisect AB, BC in E, F, and join DE, DF.

.. it must be at D, the only pt. common to ED, FD.

B

Alternative Proof.-[Draw a diameter through D.] Then if D were not the centre there could not be more than two equal st. lines drawn from D to the Oce, .. D is the centre.

[III. 7.

NOTE.

The first proof is one of Euclid's own, but is not given by Simson.

Р

PROPOSITION 10.

One circumference of a circle cannot cut another at more than two points.

If it be possible let one Oce cut another at the three pts. ABC.

Find the centre K of one of the two Os and join KA, KB,

KC.

Then rad. KA=rad. KB=rad. KC,

.. K is the centre of the other O,
which is impossible.

[III. 9.

[III. 5.

PROPOSITION 11.

If two circles touch each other internally the straight line which joins their centres, being produced, shall pass through the point of contact.

Let the two Os ABC, ADE touch each other internally at the pt. A; the st. line through their centres shall pass through A.

.. G is not the centre of ADE.

Similarly it can be shown that, if FG cut the Os at any other

pt. than A, G is not the centre of O ADE. .. the st. line joining F to centre of

ADE passes through A

when produced.

NOTE.

It follows from the above demonstration that there cannot be a second point of internal contact. For if there were a second one, K, the centre of ADE, would also lie on the line FK, which is impossible, as the two circles have not the same centre.

Ex. 280.-Two circles touch each other internally. Show that if a st. line be drawn perpr. to the diameter through the point of contact the two parts of it lying between the two circumferences are equal.

Ex. 281.-Enunciate and prove a converse of the preceding exercise.

PROPOSITION 12.

If two circles touch one another externally the straight line which joins their centres shall pass through the point of contact.

Let the two Os ABC, ADE touch each other externally at the pt. A; the join of their centres shall pass through A.

Similarly it can be shown that if FG cut the Oces at any

other pt. than A, G is not the centre,

.. the st. line joining F to the centre of ADE passes through A.

NOTE.

It follows from the above demonstration that there cannot be a second point of external contact. For if there were a second one, K, the centre of ADE, would also lie on FK produced, which is impossible. Observe the close resemblance between the above demonstration and that of III. II.

Alternative Proof.-Let two Os, whose centres are A and B, meet at a pt. C which is not on AB; they shall not touch each other externally.

Draw CP r to AB and produce it to D, so that PD=PC. Join AC, AD, BC, BD.

B

In As APC, APD

AP, PC=AP, PD,

and APC LAPD;

.. AC=AD;

.. D lies on the whose centre is A. Similarly D can be shown to lie on the circle whose centre is B,

.. CD lies in both Os;

.. they do not touch externally;

.. if two Os do touch externally they cannot meet at a pt. not on the line joining the centres.