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For, let the circle KLG be placed upon ACH, so that the centre R shall fall upon the centre 0, and the point K upon the point A: then will the chord KL

B

S coincide with AC (P. M

D IV.); and consequently,

A

K R they will be equally distant from the centre ; which was to be proved.

H

G

2°. Let AB be less than KL : then will it be at a greater distance from the centre.

For, place the circle KLG apon ACH, so that R shall fall upon 0, and K upon A. Then, because the chord KL is greater than AB, the arc KSL is greater than AMB; and consequently, the point I will fall at a point C, beyond B, and the chord KL will take the direction AC.

Draw OD and OE, respectively perpendicular to AC and AB; then will OE be greater than OF (A. 8), and OF than OD (B. I., P. XV.): hence, OE is greater than OD. But, OE and

OD are the distances of the two chords from the centre (B. I., P. XV., C. 1): hence, the less chord is at the greater distance from the centre ; which was to be proved.

Scholium. All the propositions relating to chords and arcs of equal circles, are also true for chords and arcs of one and the same circle. For, any circle may be regarded as made ap of two equal circles, so placed, that they coincide in all

their parts.

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If a straight line is perpendicular to a radius at its outer

extremity, it will be tangent to the circle at that point; conversely, if a straight line is tangent to a circle at any point, it will be perpendicular to the radius drawn to that point.

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1°. Let BD be perpendicular to the radius CA, at A : then will it be tangent to the circle at A.

For, take any other point of BD, E, and draw CE:

B

E D

O then will CE he greater than CA (B. I., P. XV.); and consequently, the point E will lie without the circle : hence, BD touches the circumference at the point A; it is, therefore, tangent to it at that point (D. 11); which was to be proved.

2o. Let BD be tangent to the circle at A: then will it be perpendicular to CA.

For, let Ebe any point of the tangent, except the point of contact, and draw CE. Then, because BD is a tangent, E lies without the circle ; and consequently, CE is greater than CA: hence, CA is shorter than any sther line that be drawn from C to BD; it is, therefore, perpendicular to BD (B. I., P. XV., C. 1); which was to be proved.

Cor. At a given point of a circumference, only one tangent can be drawn. For, if two tangents could be drawn, they would both be perpendicular to the same radius at the same point; which is impossible (B. I., P. XIV.).

PROPOSITION X.

THEOREM.

Troo parallels intercept equal arcs of a circumference,

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There may be three cases: both parallels may be secants; one may be a secant and - the other a tangent; or, both may be tangents.

Let the secants AB and DE be parallel : then will the intercepted arcs MN and PQ be equal.

For, draw the radius CH perpendicular to the chord MP; it will also be per

A

-B M

P pendicular to NQ (B. I., P.

D

N XX., C. 1), and H will be at the middle point of the arc MHP, and also of the NHQ : hence, MN, which is the difference of HŅ and IIM, is equal to PQ, which is the difference of HQ and HP (A. 3); which was to be proved.

arc

2°. Let the secant AB and tangent DE, be parallel • then will the intercepted arcs MH and PH be equal.

For, draw the radius CH to the point of contact H;

H
D-

-E it will be perpendicular to DE A

B M

P (P. IX.), and also to its parallel MP. But, because CH is perpendicular to MP, H is the middle point of the arc MIP (P: VI.) : and PH are equal ;

which was to be proved.

hence,

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3°. Let the tangents DE and IL be parallel, and let H and K be their points of contact : then will the intercepted arcs HMK and HPK be equal.

For, draw the secant AB parallel to DE; then, from

II
D-

-E what has just been shown, we

-B M

P
shall have HM equal to HP,
and MK equal to PK: hence,

Ci
HMK, which is the sum of
HM and MK, is equal to
HPK, which is the sum of I

L

K HP and PK; which was to be proved.

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If two circumferences intersect each other, the points of in

tersection will be in a perpendicular to the straight lino joining their centres, and at equal distances from it.

are

be perpen

D

Let the circumferences, whose centres C and D, intersect at the points A and B: then will CD dicular to AB, and AF will be equal to BF.

C For, the points A and B, being on the circumference

B В whose centre is C, are equally distant from C; and being on the circumference whose centre is D, they are equally dis tant from D: hence, CD is perpendicular to AB at its middle point (B. I., P. XVI., C.); which was to be proved.

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If two circumferences intersect each other, the distance be

tween their centres will be less than the sum, and greater than the difference, of their radii.

sum, and

Let the circumferences, whose centres are cand D, intersect at A: then will CD be less than the greater than the difference of the radii of the two circles.

C.

D For, draw AC and AD, forming the triangle ACD. Then will CD be less than the sum of AC and

and AD, and greater than their difference (B. I., P. VII.); which was to be proved.

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If the distance between the centres of two circles is equal

to the sum of their radii, they will be tangent externally.

Let C and D be the centres of two circles, and let the distance between the centres be equal to the sum of the radii : then will the circles be tangent externally.

For, they will have a point A, on the line CD, common, and they will have no other point in common; for, if they

C

D had two points in common, the distance between their centres would be less than the sum of their radii ; which is contrary to the hypothesis : hence, they are tangent. externally ; which was to be proved.

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