Sidebilder
PDF
ePub

PROPOSITION X.
( Argument ad absurdum).

Theorem. One circumference of a circle cannot cut another in all more than two points.

Suppose the FAB can cut O DEF in more than two points, as B, G, F, and prove, on that supposition, that k would be the centre of both Os, which (by v. 3.) is impossible, and :: the supposition is false.

PROPOSITION XI.
(Argument ad absurdum).

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

[ocr errors]

Steps of the Demonstration. Suppose that the line which joins the centres does not pass through the point of contact a, but has some other direction, as Bc; then prove, on that supposition,

1. That FG + GA > FB,
2. that GA > GB,

3. that GD > GB, i.e. less > greater, which shows

the supposition to be false, and :: that the line joining the centres must pass through A.

PROPOSITION XII.

( Argument ad absurdum). Theorem. If two circles touch each other externally, the straight line B which joins their centres shall pass through the point of contact.

Prove that the supposition, that the line joining the centres passes otherwise than through a is false, by showing that, on such supposition, the line FG would be at the same time both > and < FA + AG.

PROPOSITION XIII.

( Argument ad absurdum). Theorem. One circle cannot touch another in more points than one, whether it touches on the inside or the outside.

[ocr errors][merged small][ocr errors]

Steps of the Demonstration to Case 1st. Having supposed it possible that O EBF can touch O ABC internally in more than one point, as B and 1), show, on that supposition,

1. That right line bd falls within each o, 2. that gh passes through the point of contact,

which shows the supposition to be false.

Steps of the Demonstration to Case 2nd. Having supposed it possible that the ACK can touch O ABC externally in more than one point, as in A and c; prove, on that supposition,

1. That right line ac falls within O ACK, 2. that ac is without O ABC, 3. that ac is also within O ABC, which shows the

supposition to be false.

PROPOSITION XIV.

Theorem. Equal straight lines in a circle are equally distant from the centre; and those which are equally distant from the centre, are equal to each other.

V

riai

Steps of the Demonstration to Part 1st. 1. Prove that AF = FB, and :: AB = 2 AF, that similarly CD = 2 cG,

AF = cG,

AE- EC,
that AFP + FE' = Ego + Gc',
that

FE = EG,
that :: AB and cd are equally distant
from the centre.

that that

i20 con

Steps of the Demonstration to Part 2nd. 1. Prove that AF = co', and :. AF = cg, 2. that AB = CD.

PROPOSITION XV. Theorem. The diameter is the greatest straight line in a circle; and I of all others, that which is nearer to the centre is always greater than the K more remote; and conversely, the greater is nearer to the centre than G the less.

Part 1st. That the diameter is the greatest line in a O; and that a line nearer the centre is always > one more remote.

Steps of the Demonstration. 1. Prove that AD = BE + EC,

that AD > BC, 3. that EH < EK,

[ocr errors][ocr errors][merged small]

( Argument ad absurdum). Theorem. The straight line drawn at right angles

to the diameter of a circle from the extremity of it, falls without the circle; and no straight line can be drawn between that straight line and the circumference from the extremity, so as not to cut the circle.

Steps of the Demonstration. Part 1st. That the right line from the extremity A, at right Zs to AB, shall fall without the O.

Suppose it to fall within the O as Ac; and prove, on that supposition, 1. That each of the ZS DAC, ACD (of the A ADC)

is a right Z, which is impossible, and ::

« ForrigeFortsett »