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911. PROF. XXXV. To find the centre of a circle.

Let the circle of which it is required to find the centre be AGBF. Draw any chord AB (fig. 267.); bisect it, and from the point of division D raise a perpendicular FG: this line will pass through the centre, and consequently, if it be bisected, the point of division will be the

centre.

D

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FM

Fig. 267.

If the centre of the circle be not in the line FG, it must be somewhere out of it; for instance, at the point L. But this is impossible, for if the point L were the centre, the right line LM would be a radius; and since this line bisects the chord AB, it is (Prop. 34.) perpendicular to AB; which cannot be, since CD is perpendicular to AB.

912. PROP. XXXVI. To find the centre of an arc of a circle.

BD

Fig. 268.

Let ABDF be the are of which it is required to find the centre. Draw any two chords AB, DF (fig. 268); bisect them, and from the points of division raise the perpendiculars MC, LC; the point C, in which these two perpendiculars cut each other, is the centre of the are.

For (Prop. 35.) the perpendicular MC and the perpendicular LC both pass through the centre of the same circle; this centre must therefore be the point C, which is the only point common to the two perpendiculars.

913. PROP. XXXVII. If three equal lines meet in the same point within a circle, and are terminated, they are radii of that circle.

The lines CA, CB, CD (fig. 269.), drawn from the same point C within a circle, and terminated by it, being equal, the point C is the centre of the circle. Draw the lines AB, BD; bisect them, and let the points of division be F, G; and draw the lines CF, CG.

B

Fig. 269.

In the triangles CFA, CFB, the sides CA, CB are equal by supposition, the sides FA, FB are equal by construction, and the side CF is common. These two triangles, then, have the three sides equal; they are therefore (Prop. 5.) identical. Wherefore the two angles at Fare equal, and the line FC (Defin. 11.) is perpendicular to the chord AB. And since this perpendicular bisects the chord AB, it must (Prop. 35.) pass through the centre of the circle. In like manner, it may be demonstrated that the line GC also passes through the centre. Wherefore the point C is the centre of the circle, and CA, CB, CD are radii. 914. PROP. XXXVIII. If the radius of a circle be perpendicular to a chord, the radius bisects both the chord and the arc of the chord.

Let the radius CF be perpendicular to the chord AB (fig. 270.); the right line AD is equal to the right line DB, and the arc AF equal to the arc FB. Draw the right lines CA, CB.

D

In the large triangle ACB, the side CA (Prop. 1.) is equal to the side CB, because they are radii of the same circle. The angle A is (Prop. 4.) therefore equal to the angle B. The angles at D are right angles, and therefore equal; and the angles at C are consequently (Prop. 23.) equal. Also the side CA is equal to the side CB, and the side CD is common. These two triangles, then, having two sides and the angle contained by them equal, are (Prop. 3.) identical, whence the side AD is equal to the side DB. Again, since the angles ACF, BCF are equal, the arcs AB, BF, which measure these angles, are also equal. The chord AB and the are AFB are therefore bisected by the radius CF.

F

Fig. 270.

915. PROP. XXXIX. A right line perpendicular to the extremity of a radius is a tangent to the circle.

Let the line AB (fig. 271.) pass through the extremity of the radius CT in such a manner that the angles CTA, CTB shall be right angles; this line AB touches the circumference in only one point T. If AB touch the circumference in any other point, let it be D, and draw the line CD.

1 D

B

Fig. 271.

In the right-angled triangle CTD the square of the hypothenuse CD is equal to the two squares of CT and TD taken together. The square of CD is therefore greater than the square of CT, and consequently the line CD is greater than the line CT, which is a radius. Therefore the point D is out of the circumference. And in like manner it may be shown that every point in the line- AB is out of the circumference, except T; AB is therefore a tangent to the circle.

COROLLARY. It follows, therefore, that a perpendicular is the shortest line that can be

T

drawn from any point to a given line; since the perpendicular CT is shorter than any othe line which can be drawn from the point C to the line AB.

916. PROP. XI. If a right line be drawn touching a circumference, a radius drawn to ti point of contact will be perpendicular to the tangent.

Let the line AB (fig. 272.) touch the circumference of a circle in a point T, the radius CT is perpendicular to the tangent AB. For all other lines drawn from the point C to the line AB must pass out of the circle to arrive at this line. The line CT is therefore the shortest which can be drawn from the point C to the line AB, and consequently (Corol. to Prop. 39.) is perpendicular to the line AB.

917. PROP. XLI. The angle formed by a tangent and chord is measured by half the arc of that chord.

A

T D

B

Fig. 172.

Let BTA (fig. 273.) be a tangent and TD a chord drawn from the point of contact T the angle ATD is measured by half the arc TFD, and the angle BTD is measured b half the are TGD. Draw the radius CT to the point of contact, and the radius CF perpendicular to the chord TD.

The radius CF being perpendicular to the chord TD (Prop. 38.) bisects the are TFD. TF is therefore half the arc TFD.

B

T

Fig. 273.

In the triangle CML the angle M being a right angle, the two remaining angles are (Prop. 22.) equal to a right angle. Wherefore the angle C is that which the angle L wants of a right angle. On the other side, since the radius CT is perpendicular to the tangent BA, the angle ATD is also that which the angle L wants of a right angle. The angle ATD is therefore equal to the angle C. measured by the arc TF, consequently the angle ATD is also measured by the arc TF which is half of TFD. The angle BTD must therefore be measured by half the arc TGL since these two halves of arcs make up half the circumference.

918. PROP. XLII. An angle at the circumference of a circle is measured by half the arc by which it is subtended.

Let CTD (fig. 274.) be the angle at the circumference; it has for its measure half the are CFD by which it is subtended.

Suppose a tangent passing through the point T.

B

But the angle Ci

T

Fig. 274.

The three angles at T are measured by half the circumference (Prop. 22.), but the angle ATD is measured (Prop. 41.) by half the arc TD, and the angle BTC by half the arc TC; consequently the angle CTD must be measured by half the arc CFD, since these three halves o arcs make up half the circumference.

919. PROP. XLIII. The angle at the centre of a circle is double of the angle at the cir cumference.

Let the angle at the circumference ADB (fig. 275.) and the

angle at the centre ACB be both subtended by the same are AB,

the angle ACB is double of the angle ADB.

For the angle ACB is measured by the are AB, and the angle ADB is (Prop. 42.) measured by half the same arc AB; the angle ACB is therefore double of the angle A DB.

920. PROP. XLIV. Upon a given line, to describe a segment of

a circle containing a given angle.

Fig. 275.

Let AB (fig. 276.) be the given line and G the given angle, it is required to draw suri a circumference of a circle through the points A and B that the angle D shall be equal t the angle G

For this purpose draw the lines AL, BL in such manner that the angles A and B shall be equal to the angle G: at the extremities of LA, LB raise the perpendiculars AC, BC; and from the point C in which these two perpendiculars cut each other, with the radius CA or CB describe the circumference ADB; the angle D will be equal to the angle G.

D

The angle LAB, formed by the tangent AL and the chord AB, is (Prop. 41.) measured by half the arc AFB; and the angle D at the circumference is also measured (Prop. 42.) by half the arc AFB; the angle D is therefore equal to the angle LAB. But the angle LAB is made equal to the angle G; the angle D is therefore equa to the angle G.

Fig. 276.

921. PROP. XLV. In every triangle the greater side is opposite to the greater angle, and the greater angle to the greater side.

In the tri ngle ABC (fig. 277.), if the side AB be greater than the side AC, the angl

1

Copposite to the side AB will be greater than the angle B opposite to the side AC.
Draw the circumference of a circle through the three points A,
C, B.

Since the chord AB is greater than the chord AC, it is manifest that the are ADB is greater than the arc AFC; and consequently the angle at the circumference C, which is measured (Prop. 42.) by half the arc ADB, is greater than the angle at the circumference B, which is measured by half the arc AFC.

Again, if the angle C is greater than the angle B, the side AB opposite to the angle C will be greater than the side AC opposite to the angle B.

D

B

Fig. 277.

F

The angle C is measured (Prop. 42.) by half the arc ADB, and the angle B by half the are AFC. But the angle C is greater than the angle B; the arc ADB is therefore greater than the are AFC, and consequently the chord AB is greater than the chord AC. 922. PROP. XLVI. Two parallel chords intercept equal arcs.

If the two chords AB, CD (fig. 278.) are parallel, the arcs AC, BD are equal. Draw the right line BC.

Because the lines AB, CD are parallel, the alternate angles ABC, BCD are (Prop. 17.) equal.

But the angle at the circumference BCD is measured (Prop. 42.) by half the are AC; and the angle at the circumference BCD is measured by half the arc BD; the ares AC, BD are therefore equal.

923. PROP. XLVII. If a tangent and chord be parallel to each other, they intercept equal arcs. Let the tangent FG (fig. 279.) be parallel

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to the chord AB; the arc TA will be equal to the arc TB.

G

B

D

Fig. 279.

Draw the right line TA. Because the lines FG, AB are parallel, the alternate angles FTA, TAB are (Prop. 17.) equal. But the angle FTA, formed by a tangent and a chord, is measured (Prop. 41.) by half the arc TA, and the angle at the circumference TAB is measured (Prop. 42.) by half the arc TB. The halves of the arcs TA, TB, and consequently the arcs themselves, are therefore equal.

924. PROP. XLVIII. The angle formed by the intersection of two chords is measured by half the two arcs intercepted by the two chords.

Let the two chords AB, DF (fig. 280.) cut each other at the point C, the angle FCB or ACD is measured by half the two arcs FB, AD. Draw AG parallel to DF.

Because the lines AG, DF are parallel, the interior and exterior angles GAB, FCB are (Prop. 19.) equal. But the angle at the eircumference GAB is measured (Prop. 42.) by half the arc GFB. The angle FCB is therefore also measured by half the arc GFB.

A

G

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Fig. 280.

B

Because the chords AG, DF are par llel, the arcs GF, AD are (Prop. 46.) equal: AD may therefore be substituted in the room of GF; wherefore the angle FCB is measured by half the arcs AD, FB. 925. PROP. XLIX. The angle formed by two secants is measured by half the difference of the two intercepted arcs.

Let the angle CAB (fig. 281.) be formed by the two secants AC, AB, this angle is measured by half the difference of the two arcs GD, CB, intercepted by the two secants. Draw DF parallel to AC.

Because the lines AC, DF are parallel, the interior and exterior angles CAB, FDB are (Prop. 19.) equal. But the angle FDB is measured (Prop. 42.) by half the arc FB; the angle GAB is therefore also measured by half the arc FB.

Because the chords GC, DF are parallel, the arcs GD, CF are (Prop. 46.) equal; the arc FB is therefore the difference of the are GD and the arc CFB. Where the angle A has for its measure half the difference of the arcs GD, CFB.

G

A

D

Fig. 281.

C

F

926. PROP. L. The angle formed by two tangents is measured by half the difference of th two intercepted arcs.

Let the angle CAB (fig. 282.) be formed by the two tangents AC, AB; this angle is * measured by half the difference of the two arcs GLD, GFD. Draw DF parallel to AC. Because the lines AC, DF are parallel, the interior and exterior angles CAB, FDB are } (Prop. 19.) equal. But the angle FDB, formed by the tangent DB and the chord DF, is measured (Prop. 41.) by half the arc FD. Therefore the angle CAB is also measured by balf the arc FD.

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COROLLARY. In the same way it may be demonstrated that the angle formed by a tange ATC (fig. 283.) and a secant ADB is measured by half the difference of the two inte cepted arcs.

927. PROP. LI. To raise a perpendicular at the extremity of a given line.

At the extremity A (fig. 284.) of the given line AB let it be required to raise a pe pendicular.

From any point C taken above the line AB describe a circumference passing through the point A and cutting the line AB in any other point, as G. Draw the diameter DG and the right line AD; this line AD will be perpendicular to the line AB.

The angle DAG at the circumference is measured (Prop. 42.) by half the arc DFG, which is half the circumference, because DCG is a diameter. The angle DAG is therefore measured by one fourth

D

Fig. 294.

F

part of the circumference, and consequently (Defin. 10.) is a right angle, whence the li AD is (Prop. 11.) perpendicular to the line AB.

COROLLARY. Hence it follows that the angle at the circumference which is subtendi by a diameter must be a right angle.

928. PROP. LII. From any point without a circle to draw

a tangent to that circle.

From the point A (fig. 285.) let it be required to draw a tangent to the circle DTB.

Draw from the centre C any right line CA; bisect this right line, and from the point of division B, as a centre, describe the arc CTA. Lastly, from the point A, and through the point T, in which the two arcs cut each other, draw the right line AT; this right line AT will be a tangent to the circle DTB. Draw the radius CT.

D

F

Fig. 285.

B

The angle CTA at the circumference, being subtended by the diameter CA, is (Corol. to Prop. 51.) a right angle; therefore the line TA is perpenc cular to the extremity of the radius CT, and consequently (Prop. 40.) is a tangent to t circle DTB.

929.

SURFACES.

DEFINITIONS.-1. A mathematical point has neither length, breadth, nor thicknes The physical point, now for consideration, has a supposed length and breadth excee ingly small.

2. A physical line is a series of physical points, and consequently its breadth is equal that of the physical points whereof it is composed.

3. Since physical lines are composed of points, as numbers are composed of units, poir may be called the units of lines.

4. As to multiply one number by another is to take or repeat the first number as mai times as there are units in the second; so to multiply one line by another is to take repeat the first line as many times as there are units, that is, physical points, in t second.

930. PROP. LIII. The surface of a rectangle is equal to the product of its two sides.

Let the rectangle be ABCD (fig. 286.). If the physical line AB be multiplied by the physical line AD, the product will be the surface ABCD.

B b b b b b b b b b c

Aaa a a a a a a
Fig. 286.

If as many physical lines equal to AB as there are physical points in the line AD be raised perpendicularly upon AD, these lines AB, ab, &c. will fill up the whole surface of the rectangle ABCD. Wherefore the surface ABCD is equal to the line AB taken as many times as there are physical points in the li AD; that is, (Defin. 4.) equal to the line AB multiplied by the line AD.

931. PROP. LIV. The surface of a triangle is equal to half the product of its altitude and b If from the vertex of any angle A (fig. 287.) of the triangle BAC be drawn AD, pe

pendicular to the opposite side BC, this perpendicular is called the height, and the side BC

the base of the triangle. Now the surface of the triangle is

1 equal to half the product of the height AD and the base BC. Produce BC both ways; through the point A draw FG parallel to BC, and raise the two perpendiculars BF, CG. Because the rectangle BFGC and the triangle BAC are between the same parallels, and have the same bases, the triangle is (Prop. 29.) half the rectangle. But the surface of the rectangle is equal (Prop. 53.) to the product of BF and BC. Wherefore the surface of the triangle is equal to half the product of BF and BC, that is, of DA and GC.

932. PROP. LV. To measure the surface of any rectilineal figure.

A

Fig. 287.

Let ABCDFA (fig. 288.) be the rectilineal figure, whereof it is required to find the surface.

Divide the whole figure into triangles by drawing the lines CA, CF. Then, drawing a perpendicular from the point B to the side CA, multiply these two lines; the half of their product will (Prop. 54.) give the surface of the triangle ABC. In the same manner let the surfaces of the remaining triangles B ACF, FCD be found. These three surfaces added together will give the whole surface of the figure ABCDFA.

933. PROF. LVI. The area of a circle is equal to half the product of its radius and circumference.

If the radius of the circle C (fig. 289.) be multiplied by

its circumference, the half of the product will give the surface of the circle.

Fig. 288.

Two physical points being manifestly not sufficient to make a curve line, this must require at least three. If, therefore, all the physical points of a circumference be taken two by two, these will compose a great number of small right lines. From

the extremities L, M of one of these small right lines if two radii LC MC be drawn, a small triangle LCM will be formed, the surface of which will be equal to half the product of its height; that is, the radius and its base.

To find the surface of all the small triangles whereof the circle is composed, multiply the height, that is, the radius, by all the bases, that is, by the circumference, and take the half of the product; whence the area or surface of the circle will be equal to half the product of the radius and circumference.

934. PROP. LVII. To draw a triangle equal to a given circle.

LM

Fig. 289.

Let it be required to form a triangle the surface of which shall be equal to that of the circle AGFDA (fig. 290.).

At the extremity of any radius CA of the circle, raise a perpendicular AB equal to the circumference AGFD, and draw the right line CB. The surface of the triangle BCA will be equal to that of the circle AGFDA.

G

Fig. 290.

B

The surface of the circle is equal (Prop. 56.) to half the product of the radius CA and the circumference, or the line AB. The surface of the triangle is also equal (Prop. 54.) to half the product of its height CA, or radius, and its base BA, or circumference. Therefore the surface of the triangle is equal to that of the circle.

PROPORTION.

935. DEFINITIONS. - 1. The ratio of one quantity to another is the number of times which the first contains the second; thus the ratio of 12 to 3 is four, because 12 contains 3 four times; or, more universally, ratio is the comparative magnitude of one quantity with respect to another.

2. Four quantities are proportional, or in geometrical proportion, or two quantities are said to have the same ratio with two others, when the first contains or is contained in the second, exactly the same number of times which the third contains or is contained in the fourth; thus, the four numbers 6, 3, 8, 4 are proportionals, because 6 contains 3 as many times as 8 contains 4, and 3 is contained in 6 as many times as 4 is contained in 8, that is, twice; which is thus expressed: 6 is to 3 as 8 to 4; or 3 is to 6 as 4 to 8. 936. PROP. LVIII Parallelograms which are between the same parallels are to one a

other as their buses.

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