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(2.) If from a point without a circle, two straight lines be drawn to the concave part of the circumference, making equal angles with the line joining the same point and the centre, the parts of the lines which are intercepted within the circle are equal.

From the point P without the circle ABC let two lines PB, PD be drawn making equal angles with PO, the line joining P and the centre; AB shall be equal to CD.

E

B

P

F

Let fall the perpendiculars OE, OF; then since the angle at E is equal to the angle at F, and EPO-FPO, and the side PO, opposite to one of the equal angles in each is common, .. OE OF, and consequently (Eucl. iii. 14.) AB=CD.

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(3.) Of all straight lines which can be drawn from two given points to meet on the convex circumference of a given circle; the sum of those two will be the least, which make equal angles with the tangent at the point of concourse.

Let A and B be two given points, CE a tangent to the circle at C, where the lines AC, BC make equal angles with it; and let lines AD, BD be drawn from A and B to any other point D on the

E

B

convex circumference; AC and CB together are less than AD, DB together.

D

Let AD meet the tangent in E. Join EB, then (i. 6.) AC and CB together are less than AE and EB; but AE, EB are less than AD, DB (Eucł. i. 19.), .. a fortiori AC, CB are less than AD, DB. And the same may be proved of lines drawn to every other point in the convex circumference.

(4.) If a circle be described on the radius of another circle; any straight line drawn from the point where they meet, to the outer circumference, is bisected by the interior one.

Let ADB be a circle described on the radius AB of the circle ACE. Draw any line AC meeting the circle ABD in D; AD is equal to DC.

Join DB. Then the angle ADB

being in a semicircle is a right angle;

E

B

and therefore BD being drawn from the centre B of the circle ACE bisects AC (Eucl. iii. 3.).

(5.) If two circles cut each other, and from either point of intersection diameters be drawn; the extremities of these diameters and the other point of intersection shall be in the same straight line.

Let the two circles ABC, ABD cut each other in A and B, draw the diameters AC, AD, and join BC, BD; CB and BD are in the same straight line.

Join AB; the angles ABC, ABD being angles in semicircles are right angles, and therefore (Eucl. i. 13.) CB and BD are in the same straight line.

(6.) If two circles cut each other, the straight line joining their two points of intersection, is bisected at right angles by the straight line joining their centres.

Let the two circles whose centres are C and D cut each other in A and B; join AB, DC. bisects AB at right angles.

DC

Join BD, DĄ, AC, CB.

D

B

Since AD=DB and DC is common to the triangles ADC, BDC and the base AC= CB, .. (Eucl. i. 8.) the angle ADE = BDE. Hence the two sides AD, DE are equal to the two BD, DE and the included angles are equal, .. (Eucl. i. 4.) AE = EB, and the angle DEA=DEB, and being adjacent, they are right angles, i. e. DC bisects AB at right angles.

(7.) To draw a straight line which shall touch two given circles.

1. If the circles be equal.

Let A and B be the centres, join AB; and from A

and B draw AC, BD at right angles to it; join CD. Then AC being parallel and equal to DB; CD is parallel to AB, :. CABD is a rectangular parallelogram; and the angles at C and D being right angles, CD is a tangent to both circles (Eucl. iii. 16. Cor.).

2. If the circles be unequal, and the line be required to touch them on the same side of the line joining the

centres.

Let A and B be the centres; join AB; and with the

centre B and distance equal to the difference of the given radii, describe a circle, and from A draw AE touching it. Join BE and produce it to D, draw AC parallel to BD, and join CD.

Then AC being parallel and equal to DE, CD is equal and parallel to AE, :. ACDE is a parallelogram ; and the angle AEB being a right angle, AED is also a right angle; hence the angles at C and D are right angles, and therefore CD touches both circles.

3. If the line be required to touch them on opposite sides of the line joining the centres.

With the centre B and radius equal to the sum of

E

the given radii describe a circle, to which from A draw a tangent AE. Join BE, and let it cut the given circle in D. Draw AC parallel to BE; join CD.

Then AC being equal and parallel to ED, ACDE is a parallelogram, and the angle AED being a right angle, the angles at C and D are right angles, and therefore CD touches both circles.

(8.) If a line touching two circles cut another line joining their centres, the segments of the latter will be to each other as the diameters of the circles.

Let the line AB touch the circles, whose centres are C and D, in A and B, and cut CD in

the point E; CE will be to

ED in the ratio of the diameters of the circles.

E

B

Join CA, BD. Then the angles at A and B are right angles and the angles at E are vertically opposite, therefore the triangles AEC, BED are equiangular, and consequently

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(9.) If a straight line touch the interior of two concentric circles, and be placed in the outer; it will be bisected at the point of contact.

Let AB touch the interior of two circles, whose common centre is 0, in the point C; AB is bisected in C.

2

Join OC; then (Eucl. iii. 18.) the

angles at Care right angles; and OC drawn

D

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from the centre of the circle ADB at right angles to AB,

bise cts it (Eucl. iii. 3.).

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