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9. If the point of a pair of compasses be applied to the point G (fig. 225.), and a cir

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cumference NRB be described, the are NR contained within the two lines GL, GM will measure the magnitude of the angle LGM. If the arc NR, for example, be an arc of 40 degrees, the angle LGM is an angle of 40 degrees. There are three kinds of angles (fig. 226.): a right angle (I), which is an angle of 90 degrees; an obtuse angle (II), which contains

more than 90 degrees; and an acute angle N
(III), which contains less than 90 degrees. I II
One line is perpendicular to another when - Ili
the two angles it makes with that other

line are equal: thus, the line CD (fig.
227.) is perpendicular to the line AB, if
the angles CDA, CDB contain an equal number of degrees.

Fig. 226.

. Two lines are parallel when all perpendiculars drawn from one to the other are equal;

thus, the lines FG, A B (fig. 228.) are pa- rallel, if all the perpendiculars ca, ca, &c. c * * : *4444 a are equal. i | A triangle is a surface enclosed by three ; ! right lines, called sides (fig. 229.). An equilateral triangle (I) is that which has three sides equal; an isosceles triangle has | only two of its sides equal (II); a scalene lo +. triangle (III) has its three sides unequal. Fig. 227. Fig. 228. A quadrilateral figure is a surface enclosed by four right limes, which are called its sides. A parallelogram is a quadrilateral figure, which has its opposite sides parallel; thus,

Fig. 229 Fig. 230. if the side BC (fig. 230.) is parallel to the side AD, and the side AB to the side DC, the quadrilateral figure ABCD is called a parallelogram. A rectangle is a quadrilateral figure all the angles p C whereof are right angles, as ABCD (fig. 231.). A square is a quadrilateral figure whose sides are all equal and its angles right angles (fig. 232.). A b

A trapezium is any quadrilateral figure not a Fig. 231. Fig. 252. parallelogram.

. Those figures are equal which enclose an equal space; thus, a circle and a triangle are

equal, if the space included within the circumference of the A d circle be equal to that contained in the triangle. Those figures are identical which are equal in all their parts; that is, which have all their angles equal and their sides equal, and enclose equal spaces, as BAC, EDG (fig. 23.3.). It is manifest that two figures are identical which, being placed one upon the other, perfectly coincide, for in that case they must be equal in all their parts. It must be ob- C e G ol. that a line merely so expressed always denotes a right Fig. 235,

line.

Axiom. Two right lines cannot enclose a space; that category requires at least three lines.

876. Proposition I. The radii of the same circle are all equal.

The revolution of the line AB about the point A (fig. 234.) being necessary (Defin. 5.) to form the circle BCDFGLB, when in revolving the point B is upon the point C, the whole line AB must be upon the line AC; otherwise two right lines would enclose a space, which is impossible: wherefore the radius AC is equal to the radius AB. In like manner it may be proved that the radii AB, AF, AG, &c. are all equal to AB, and are therefore equal among themselves.

877. Phor. II. On a given line to describe an equilateral tri. angle.

RIGHT LINES AND REctilin EAL Figu R.E.S.

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Let AB (fig. 235.) be the given line upon which it is required to describe a triangle

whose three sides shall be equal. i) C r From the point A, with the radius AB, describe the circumference BCD, and from the point B, with the radius BA, de- ^ scribe the circumference ACF; and from the point C, where these two circumferences cut each other, draw the two right lines CA, CB. Then ACB is an equilateral triangle.

For the line AC is equal to the line AB, because these two lines are radii of the same circle BCD ; and the line BC is Fig. 235 equal to the line A B, because these two lines (Prop. 1.) are radii of the same circle ACF. Wherefore the lines AC and BC, being each equal to the line AB, are equal to one another, and all the three sides of the triangle ACB are equal; that is, the triangle is equilateral.

878. Prop. III. Triangles which have two sides and the angle subtended or contained by them equal are identical.

In the two triangles BAC, FDG (fig. 236.), if the side DF be equal to the side AB, and the side DG equal to the side AC, and also the angle at D

equal to the angle at A, the two triangles are identical. A D Suppose the triangle FDG placed upon the triangle BAC in such manner that the side DF fall exactly upon the side equal to it, A.B. Since the angle D is equal to the angle A, the side DG must fall upon the side equal to it, AC; also the point F B C r G

will be found upon the point B, and the point G upon the point C: consequently the line FG must fall wholly upon the line BC, otherwise two right lines would enclose a space, which is impossible. Wherefore the three sides of the triangle FDG coincide Fig.256 in all points with the three sides of the triangle BAC, and the two triangles have their sides and angles equal, and enclose an equal space; that is (Defin. 20.), they are identical. 879. Pror. IV. In an isosceles triangle the angles at the base are equal. Let the triangle BAC (fig. 237.) have its sides AB, AC equal, the angles B and C at the base are also equal. Conceive the angle A to be bisected by the right line AD. In the triangles BAD, DAC the sides AB, AC are, by sup- B position, equal; the side AD is common to the two triangles, and the angles at A are supposed equal. These two triangles, therefore, have two sides, and the angle contained by them equal. Hence, they are identical (Prop. 3), or have all ther parts equal: whence the angles B and C must be equal. 880. Pror. V. Triangles which have their three sides equal are identical. A In the two triangles ACB, FDG (fig. 238.), let the side AC be equal to the side FD, the side CB equal to the side DG, and the side A B to the side FG ; these two triangles are identical. C ld Let the two triangles be so joined that the side FG shall coincide with the side A B (fig. 239.), and draw the right line CD. Since in the triangle CAD he G the side AC is equal to the side AD, Fig. 258. Fig. 459. the triangle is isoceles; whence (Defin. 13.) the angles m and n at the base are equal. Since in the triangle CBD the side BC is equal to the side BD, the triangle is isosceles; whence (Defin. 13.) the angles r and s at the base are equal. D Because the angle m is equal to the angle n, and the angle r equal to the angle s, the whole angle C is equal to the whole

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Fig. 237.

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angle D. Lastly, because in the two triangles ACB, ADB the side AC is equal to the side AD and the side CB equal to the side DB, A B

also the angle C equal to the angle D, these two triangles have two
sides, and the contained angle equal, and are therefore (Prop. 3.)
identical.
881. Prop. VI. To diride a right line into two equal parts.
Let the right line which it is required to divide into two equal F.
parts be . A B (fig. 240.). Upon AB draw (Prop. 2.) the equi- Fig. 210.
lateral triangle ADB, and on the other side of the same line

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AB draw the equilateral triangle AFB, draw also the right line DF; AC is equal to CB. In the two larger triangles DAF, DBF the sides DA, DB are equal, because they are the sides of an equilateral triangle; the sides AF, BF are equal for the same reason; and the side DF is common to the two triangles. These two triangles, then, have their sides equal, and consequently (Prop. 5.) are identical, or have all their parts equal; wherefore the two angles at D are equal. Again, in the two smaller triangles ADC, CDB the side DA is made equal to the side DB, and the side DC is common to the two triangles; also the two angles at D are equal. Thus these two triangles have two sides and the contained angle equal; they are therefore (Prop. 3.) identical, and AC is equal to CB; that is, AB is bisected. 882. Prof. VII. From a given point out of a right line to draw a perpendicular to that line. Let C (fig. 241.) be the point from which it is required to draw a perpendicular to the rght line AB. From the point C describe an arc of a circle which shall cut the line AB in two points F and G. Then bisect the line FG, ," | *, and to D, the point of division, draw the line CD : this line is ,”

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mon. These two triangles, then, having the three sides equal, are identical (Prop. 5.).

Whence (Defin. 20.) the angle CDA is equal to the angle CDB, and consequently (Defin.

11.) the line CD is perpendicular to the line AB. 883. Phor. VIII. From a given point in a right line to raise a perpendicular upon that

line. From the point C (fig. 242.), let it be required to raise a perpendicular upon the right line A.B. id

In AB take at pleasure CF equal to CG ; upon the line FG describe an equilateral triangle FDG, and draw the line CD; this line will be perpendicular to AB.

In the triangles FDC, CDG the sides DF, DG are equal, because they are the sides of an equilateral triangle; the sides FC, A s—a tiCG are equal by construction; and the side DC is common. rose These two triangles, then, having the three sides equal, are (Prop. g. z+z. 5.) identical. Therefore (Defin. 20.) the angle DCA is equal to the angle DCB, and consequently (Defin. 11.) the line CD is perpendicular to the line AB.

884. Pror. IX. The diameter of a circle dirides the circumference into two equal

b

s. ro, ADBLA (fig. 243.) be a circle; the diameter ACB bisects the circumference, that is, the arc ALB is equal to the arc ADB. G Conceive the circle to be divided, and the lower segment ..~". ACBLA to be placed upon the upper ACBDA; all the points of the arc ALB will fall exactly upon the arc ADB; and consequently these two arcs will be equal. For if the point L, for instance, does not fall upon the arc ADB, it must fall either above this arc, as at G, or below it, as at F. If it fall on G, the radius CL will be greater than the radius CD; if it falls on F, the radius CL will be less than the radius CD, I, which is (Prop. 1.) impossible. The point L, then, must fall upon Fig. 243. the arc ADB. In like manner it may be proved that all the other points of the arc ALB must fall upon the arc ADB : those two arcs are therefore equal. 885. Paor. X. A right line which meets another right line forms with it two angles, which, together, are equal to two right angles. The line AC (fig.244.) meeting the line DF, and forming with it the two angles ACD, ACF, these two angles are together equal to two right angles. From the point C as a centre describe at pleasure a circumference. NGLMN. The line NCL, being a diameter, divides the circumference (Prop. 9.) into two equal parts. The arc NGL is therefore Fig. 244. half the circumference, which contains (Defin. 6.) 180, or twice 90 degrees. Therefore the angles ACD, ACF, which, taken together, are measured by the arc NGL, are twice 90 degrees, that is (Defin. 10.), are equal to two right angles. 886. Paor, XI. A line drawn perpendicularly to another right line makes right angles with it.

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If the line CD (fig. 245.) be perpendicular to the line AB, the angle CDA is a right angle, and also the angle CDB.

For the line CD, meeting the line AB, forms c F with it two angles, which are together (Prop. G.------ • 12 10.) equal to two right angles; and these two o > angles are equal, because CD is perpendicular ; S : to A B. Wherefore each angle is a right angle. So c so

887. Prop. XII. If two lines cut each other, -i > ----- .* the rertical or opposite angles are equal. A. D b

Let the lines AB, D F (fig. 246.) cut each Fig. 245. Fig. 246.

other at the point C; the angles ACD, FCB, which are called vertical or opposite angles, are equal. From the point C, as a centre, describe at pleasure a circumference NGLMN. Since the line NCL is a diameter, the arc NGL is (Prop. 9.) half the circumference; therefore the arcs NGL, GLM are equal. From these two arcs take away the common part GL, there will remain the arc NG equal to the arc LM. Consequently the angles ACD, FCB, which are measured by these two arcs, are also equal. 888. Pror. XIII. If a line be perpendicular to one of two parallel lines, it is also pericular to the other. Let AB, CD (fig. 247.) be two parallel lines: if the line FG makes right angles with CD, it will also make right angles with A.B. Take at pleasure GC equal to GD; at the points C and D raise the perpendiculars CA, DB, and draw the lines GA, G.B. In the two triangles ACG, BDG, because the line AB is parallel to the line CD, the perpendiculars CA, D B are necessarily equal, as appears from the definition of parallel lines (Defin. 12.); the lines CG, DG are equal by construction; and the angles C and D are right angles. The two triangles ACG, BDG have then two sides and the contained angle equal, they are therefore (Prop. 3.) identical. Whence the side GA is equal to the side GB, and the angle m equal to the angle n. Again, in the triangles AGF, FGB the side GA is equal to the side GB, as has been proved, and the side GF is common. Moreover, the angle r is equal to the angle s; for if from the two right angles FGC, FG D be taken away the equal angles m and n, there will remain the equal angles r and s. The triangles AGF, FGB have then two sides and the contained angle equal; they are therefore (Prop. 3.) identical.

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Wherefore the angles GFA, GFB are equal, and consequently A. Tso o are right angles. S. 889. Prop. XIV. If one line be perpendicular to two other lines, th these two lines are parallel. Let the line FG (fig. 248.) make right angles with the lines AB and CD; these two lines are parallel. C. d

If the line AB be not parallel to the line CD, another line, ToTT as NH, may be drawn through the point F, parallel to the line Fig. 248. CD. But this is impossible; for if the line N H were parallel to the line CD, the line FG making right angles with CD would also (Prop. 13.) make right angles with NH 3 which cannot be, because, by supposition, it makes right angles with AB. 890. Paor, XV. The opposite sides of a rectangle are parallel. p C In the rectangle ABCD (fig. 249.) the side BC is parallel to - -the side AD, and the side AB parallel to the side DC. Produce each of the sides both ways. The line AB is perpendicular to the two lines BC, AD; the two lines BC, AD are therefore (Prop. 14.) parallel. In like manner, the line AD is perpendicular to the two lines AB, DC; --the two lines AB, DC are therefore (Prop. 14.) parallel. A D 891. Pror. XVI. The opposite sides of a rectangle are equal. In the rectangle ABCD (see fig. 249.) the side AB is equal to the side DC, and the side BC equal to the side AD. For, since the side BC is parallel to the side AD, the perpendiculars AB, DC are (Defin. 12.) equal; and since the side AB is parallel to the side DC, the perpendiculars BC, AD are equal. 892. Pror. XVII. A right line falling upon parallel lines makes the alternate angles equal. to, the line FG (fig. 250.) cut the parallels AB, G D ; the angles AFG, FGD, which are called alternate angles, are equal. From the point G draw GL perpendicular to the line AB, and from the point F draw FM perpendicular to the line G.D. Since the line GL is perpendicular to AB, it is also (Prop. 13.) perpendicular to the

Fig. 249.

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parallel line A.B. Whence the quadrilateral figure GLFM is a rectangle, its four angles being right angles. In the triangles GLF, FMG the sides LF, GM are equal, because they are opposite sides of the same rectangle; the sides LG, FM are equal for the same reason; and the side FG is common. The two triangles GLF, FMG have then the three sides equal, and consequently (Prop. 5.) are identical. Wherefore the angle LFG opposite to the side LG is equal to the angle FGM opposite to the side FM, Remark. In identical triangles the equal Fig. 250. Fig. 251. angles are always opposite to equal sides, as by this proposition appears. 893. PRor. XVIII. If one right line falling upon two others makes the alternate angles equal, these two lines are parallel. Let the alternate angles AFG, FG D (fig. 251.) be equal; the lines AB, G D are parallel. If the line AB is not parallel to the line GD, another line, as NH, may be drawn through the point F parallel to G.D. But this is impossible; for if the line NH were parallel to the line GD, the angle FGD would be (Prop. 17.) equal to the angles NFG, since these two angles would be alternate angles between two parallel lines; which cannot be, because, by supposition, the angle FG D is equal to the angle AFG. 894. Prop. XIX. If one right line falls upon two parallel right lines, it makes the interior angle equal to the erterior. Let the line FG (fig. 252.) meet the parallel lines BA, DC, the interior angle r is equal to the exterior angle z. Produce the lines B.A., D.C. The angle r (Prop. 17.) is equal to the angle s, because these are alternate angles, made by a right line falling upon two * * parallel lines, and the angles s and z are (Prop. 12.) equal, because they are vertical or opposite angles; therefore the angle r is equal to the angle z. 895. Pnor. XX. If one right line falling upon two other right lines makes the internal angle equal to the earternal, those two lines are parallel. Let the internal angle r (fig. 253.) be equal to the external angle z, the lines B.A, DC are parallel. The angle r is equal to the angle z by supposition, and the angle z (Prop. 12.) is equal to the angle s, because they are opposite angles. The alternate angles r, s are therefore equal, and consequently (Prop. 18.) the lines B.A., DC are parallel. 896. PRop. XXI. Through a given point to draw a line parallel A to a given line. / Let G be the point through which it is required to draw a line Fig. 255. parallel to the given line M.F. From any point G (fig. 254.) describe, at pleasure, the arc FN ; from the point F, in which the arc FN cuts the line MF, with the distance G F describe the arc GM meeting the line MF in M ; then make FL equal to GM, and draw the line GL; this line is parallel to the line M.F. Draw the line G F. The arcs GM, FL are equal by construction; therefore the alternate angles r, s, which are measured by these arcs (Defin. 9.), are equal; and consequently (Prop. 18.) the lines

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GL, MF are parallel. Fig. 25t. Fig. 255. 897. Prop. XXII. The three angles of a triangle taken together are equal to two right angles. In the triangle BAC (fig. 255.), the three angles B, A, C are together equal to two right angles.

Produce the side BC both ways; through the point A draw a line FG parallel to BC; and from the point A, as a centre, describe any circumference LMN. The angle B (Prop. 17.) is equal to the angle a, because these are alternate angles made by a right line falling upon two parallel lines. For the same reason the angle C is equal to the angle y. Because LAN is a diameter, the arc LMN is half the circumference; therefore the three angles r, A, y, which are measured by this arc, are together equal to two right angles.

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