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

For the line CD, meeting the line AB, forms with it two angles, which are together (Prop. 10) equal to two right angles; and these two angles are equal, because CD is perpendicular to AB. Wherefore each angle is a right angle. 887. PROP. XII. If two lines cut each other, the vertical or opposite angles are equal.

C

B

Fig. 245.

Fig. 246.

Let the lines AD, BF, (fig. 246.) cut each other at the point C; the angles ACB, FCD, which are called vertical er opposite angl 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 circumferenc therefore the arcs NGL, GLM are equal. From these two arcs take away the comm part GL, there will rema n the arc NG equal to the arc LM. Consequently the ang] ACB, FCD, which are measured by these two arcs, are also equal.

888. PROP. XIII. If a line be perpendicular to one of two parallel lines, it is also p pendicular to the other.

Let AB, CD (fig. 247.) be two parallel lines if the line FG makes right angles wi CD), it will also make right angles with AB.

Take at pleasure GC equal to GD; at the points C and D raise the perpendiculars CA, DB, and draw the lines GA, GB.

In the two triangles ACG, BDG, because the line AB is parallel to the line CD, the perpendiculars CA, DB 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.

A

F

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G

Fig 247.

B

Again, in the triangles AGF, FGB the side GA is equal to the side GB, as has be proved, and the side GF is common. Moreover, the angle r is equal to the angle s; if from the two right angles FGC, FGD be taken away the equal angles m and n, the will remain the equal angles r and s. The triangles AGF, FGB have then two sides a the contained angle equal; they are therefore (Prop. 3.) identical. Wherefore the angles GFA, GFB are equal, and consequently are right angles.

889. PROP. XIV. If one line be perpendicular to two other lines, 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.

If the line AB be not parallel to the line CD, another line,

as NH, may be drawn through the point F, parallel to the line CD.

A

N

G

P

Fig. 248.

B

But this is impossible; for if the line NH were parallel to the line CD, the li FG making right angles with CD would also (Prop. 13.) make right angles with NI which cannot be, because, by supposition, it makes right angles with AB. 890. PROP. XV. The opposite sides of a rectangle are parallel. In the rectangle ABCD (fig. 249.) the side BC is parallel to the side AD, and the side AB parallel to the side DC. cach of the sides both ways.

Produce

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.

891. PROP. XVI. The opposite sides of a rectangle are equal. In the rectangle ABCD (see fig. 249.) the side AB is equal

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to the side DC, and the side BC equal to the side AD. For, since the side BC is paral

to the side AD, the perpendiculars AB, DC are (Defin. 12.) equal; and since the si AB is parallel to the side DC, the perpendiculars BC, AD are equal.

892. PROP. XVII. A right line falling upon parallel lines makes the alternate any equal.

Let the line FG (fig. 250.) cut the parallels AB, GD; the angles AFG, FGD, whi are called alternate angles, are equal. From the point G draw GL perpendicular to line AB, and from the point F draw FM perpendicular to the line GD.

Since the line GL is perpendicular to AB, it is also (Prop. 19.) perpendicular to

parallel line GD Whence the quadrilateral figure GLFM is a rectangle, its four angles being right angles.

des of the same

A L

В

A

In the triangles GLF, FMG the sides LF, GM are equal, because they are opposite 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

M D

G

D

Fig. 250.

Fig. 251.

angles are always opposite to equal sides, as by this proposition appears.

893. PROP. XVIII. If one right line fulling upon two others makes the alternate angles qual, these two lines are parallel.

Let the alternate angles AFG, FGD (fig. 251.) be equal; the lines AB, GD 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 GD. 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 angle NFG, since these two angles would be alternate angles between two parallel lines; which cannot be, because, by supposition, the angle FGD is equal to the angle AFG.

894. PROP. XIX. If one right line falls upon two parallel right nes, it makes the interior angle equal to the exterior.

Let the line FG (fig. 252.) meet the parallel lines BA, DC, the interior argler is equal to the exterior angle z.

the lines BA, DC.

Produce

The angler (Prop. 17.) is equal to the angle s, because these re alternate angles, made by a right line falling upon two parallel lines, and the angles s and z are (Prop. 12.) equal, be

F

B

D

Fig. 252.

G

cause they are vertical or opposite angles; therefore the angle r is equal to the angle z. 895. PROP. XX. If one right line falling upon two other right lines makes the internal

angle equal to the external, those two lines are parallel.

Let the internal angler (fig. 253.) be equal to the external

ngle 2, the lines BA, DC are parallel.

The angle is equal to the angle z by supposition, and the

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From any point G (fig. 254.) describe, at pleasure, the are FN; from the point F, in which the are FN cuts the line MF. with the distance GF describe the arc GM meeting

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7. PROP. XXII. The three angles of a triangle taken together are equal to two right

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; id from the point A, as a centre, describe any circumference L.MN.

The angle B (Prop. 17.) is equal to the angler, because these are alternate angles made by a right line falling upon two parallel lines. to the y. For the same reason the angle C is equal Because LAN is a diameter, the arc LMN is half the circumference; therefore the are angles z, A, y, which are measured by this are, are together equal to two right angles

angle

But the angle is equal to the alternate angle B, and the angle y to the altern angle C.

Therefore, substituting B for x, and C for y, the three angles B, A, C are together eq to two right angles. COROLLARY. Hence, if two angles of any triangle be known, the third is also foun since the third angle is that which the other two taken together want of two rig angles.

898. PROP. XXIII. If two triangles have two angles equal, they have also the third an equal.

In the two triangles BAC, FDG (fig. 256.), if the angle B is equal to the angle F, and the angle A equal to the angle D, the angle C will also be equal to the angle G.

Since the angle C (Corol. to Prop. 22.) is that which the angles B and A together want of two right angles; and since the angle G is that which F and D together want of two right angles; the angles B and A being equal to the angles F and D, the angle C B must be equal to the angle G.

A

C F
Fig. 256.

899. PROP. XXIV. The exterior angle of any triangle is equal to the two interior a opposite angles taken together.

In the triangle BAC (fig. 257.) produce one of the sides BC; the angle ACD, which is called exterior, is equal to the two interior and opposite angles B and A taken together.

The line AC meeting the line BD forms with it two angles, which are together (Prop. 10.) equal to two right angles; the angle ACB is therefore that which the angle ACD wants of two right angles. But the angle ACB is (Corol. to Prop. 22.) also that which the angles B and A together want of two right angles. Wherefore the angle ACD is equal to the two angles B and A taken together.

F Li

A

M

B

C

Fig. 237.

900. PROP. XXV. Triangles which have two angles and the side which lies between th equal are identical.

In the two triangles BAC, FDG (fig. 258.), if the angle F is equal to the angle B, t angle G equal to the angle C, and the side FG equal to the side BC, these two triangles are identical.

Conceive the triangle FDG placed upon the triangle BAC in such a manner that the side FG shall fall exactly upon the equal side BC. Since the angle F is equal to the angle B, the side FD must fall upon the side BA; and since the angle G is equal to the angle C, the side GD must fall upon the side CA. Thus the three sides of the triangle FDG will be exactly placed upon the three sides of the triangle BAC; and consequently the two triangles (Prop. 5.) are identical.

1

B

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G

Fig. 258.

901. PROP. XXVI. If two angles of a triangle are equal, the sides opposite to the angles are also equal.

A

Conceive the angle A (fig. 259.) to be bisected by the line A D.

In the triangles BAD, DAC the angle B is equal to the angle C by supposition, and the angles at A are also equal. These two triangles have their two angles equal; the third angle will therefore (Prop. 23.) be equal; whence the angles at D are equal. Moreover, the side AD is common to the two triangles. These two triangles, therefore, having two angles and the side which lies between them equal, are (Prop. 25.) identical. to the side AC.

B

D

Fig. 2.59. Wherefore the side AB is equ

902. PROP. XXVII. The opposite sides of a parallelogram are equal. In the parallelogram ABCD (fig. 260.), the side AB is equal to the side DC, and the side BC equal to the side AD. Draw the line BD, which is called the diagonal. Because BC is parallel to AD, the alternate angles m and n are equal. In like manner, because AB is parallel to DC, the alternate angles r and s are equal. to the two triangles BAD, BCD. two angles and the side which lies between them equal, and are therefore (Prop. 3.) identical.

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posite to the angle n is (Prop. 26.) equal to the side DC opposite to the angle and the side BC opposite to the angle s is equal to the side AD opposite to the equ angle r

COROLLARY. Hence, the diagonal bisects the parallelogram; for the triangles BAD, BCD, having the three sides equal, are identical.

03. PROP. XXVIII.

Bane base, are equal.

Parallelograms which are between the same parallels, and have the

BCFG

Let the two parallelograms ABCD, AFGD (fig. 261.), be between the same parallels BG, AM, and upon the same base AD; the space enclosed within the parallelogram ABCD is equal to the space enclosed within the parallelogram AFGD.

In the two triangles BAF, CDG the side BA of the former triangle is equal to the side CD of the latter, because they are opposite sides of the same parallelogram. For the same reason, the side FA is equal to the side GD. Moreover, BC is equal to AD, because they are opposite sides of the same parallelogram. For the same reason, AD is equal to FG.

BC is therefore

A

D

M

Fig. 261.

equal to FG. If to both these CF be added, BF will be equal to CG. Whence the two triangles BAF, CDG, having the three sides equal, are (Prop. 5.) identical, and consequently have equal surfaces.

If from these two equal surfaces be taken the small triangle CLF, which is common, there will remain the trapezium ABCL, equal to the trapezium LFGD. To these two trapezia add the triangle ALD, and the parallelogram ABCD will be equal to the parallelogram AFGD.

904. PROP. XXIX. If a triangle and a parallelogram are upon the same base, and parallels, the triangle is equal to half the paral

between the same

lelogram.

let the parallelogram ABCD (fig. 262.) and the triangle AFD he upon the same base AD, and between the same parallels BG, AL; the triangle AFD is half the parallelogram ABCD. Draw DG parallel to AF.

Because the parallelogram AFGD is bisected by the diagonal FD (Prop. 27. Corol.), the triangle AFD is half the parallelogram AFGD. But the parallelogram AFGD is equal to

B

C F

G

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the parallelogram ABCD, because these two parallelograms are upon the same base, and parallels; therefore the triangle AFD is equal to half the parallelogram

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905. PROP. XXX. Parallelograms which are between the same parallels, and have equal bases, are equal.

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Because AD is equal to LM, and LM to FG, AD is equal to FG; and they are parallel by construction. Also AF and DG are parallel; for if DG be not parallel to AF, another line may be drawn parallel to it; whence FG will become greater or less than AD. AF and DG are therefore parallel, and AFGD a parallelo

gram.

Now the parallelogram ABCD is (Prop. 28.) equal to the parallelogram AFGD, because these two parallelograms are between the same parallels, and have the same base AD. And the parallelogram AFGD is equal to the parallelogram LFGM, because these parallelograms are between the same parallels, and have the same base FG. The parallelogram ABCD is therefore equal to the parallelogram LFGM.

two

06. PROP. XXXI. Triangles which are between the same parallels, and have equal bases, are equal.

Let the two triangles ABD, LFM (see fig. to preceding Proposition) be between the same parallels BG, AM, and upon the equal bases AD, LM; these two triangles are equal. Draw DC parallel to AB, and MG parallel to LF.

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parallelograms ABCD, LFGM are equal (Prop. 30.), because they are between parallels, and have equal bases. But the triangle ABD is (Prop. 29.) one half of the parallelogram ABCD, and the triangle LFM is one half of the parallelogram LFGM; triangles are therefore equal.

these two

907. PROP. XXXII. In a right-angled triangle, the square of the hypotenuse, or side saltending the right angle, is equal to the squares of the sides which contain the right

angle.

In the triangle BAC (fig. 264.), let the angle A be a right angle. Upon the hypotenuse BC describe the square BDFC; upon the side AB describe the square ALMB, and upon the side AC the square ARNC; the square BDFC is equal to the two squares ALMB, ARNC taken together.

Draw the right lines MC, AD, and draw AG parallel to BD.

Because the square or parallelogram MLAB and the triangle MCB are between the same parallels LC, MB, and have the same base MB, the triangle MCB is (Prop. 29.) equal to half the square ALMB.

Again, because the rectangle or parallelogram DGPB and the triangle DAB are between the same parallels GA and DB, and have the same base DB, the triangle DAB is (Prop. 29.) equal to half the rectangle DGBP.

M

Fig. 264.

F

C

N

Further, since the side MB of the triangle MBC and the side AB of the triangle ABD are sides of the same square, they are (Defin. 17.) equal. Also, since the side BC of the first triangle and the side BD of the second triangle are sides of the same square, they ar equal. And because the angle MBC of the first triangle is composed of a right angle and the angle x, and the angle ABD of the second triangle is composed of a right angle and the same angler, therefore these two angles, contained between the equal sides MB, BO and AB, BD, are equal. Wherefore the two triangles MBC, ABD, having two sides and the contained angle equal, are (Prop. 3.) identical, and consequently equal.

But the triangle MBC is half the square MLAB, and the triangle ABD is half th rectangle BDGP; the square and the rectangle are therefore equal.

In the same manner it may be demonstrated that the square ARNC and the rectangl CFGP are equal. Wherefore it follows that the whole square BDFC is equal to the two squares MLAB, ARNC taken together.

CIRCLES.

908. DEFINITIONS.-1. A right line (fig. Prop. 33. AB) terminated both ways by th circumference of a circle is called a chord.

2. A line (fig. Prop. 39. AB) which meets the circumference in one point only is calle a tangent; and the point T is called the point of contact.

3. An angle (fig. Prop. 33. ABD) which has its vertex in the circumference of a circl is called an angle in the circle.

4. A part of a circle confined between two radii (fig. Prop. 34. ACBFA) is called a sector 5. A part of a circle (fig. Prop. 35. AGBDA) terminated by a chord is called a segmen of a circle.

909. PROP. XXXIII. To draw the circumference of a circle through three given points. Let there be three given points, A, B, D (fig. 265.), through which it is required to draw the circumference of a circle. Draw the right lines AB, BD, and bisect them: from the points of the division F, G, raise the perpendiculars BC, GC; and at the point C with the radius CA describe the circumference of a circle; this circumference will pass through the points B and D. Draw the lines CA, CB, CD.

In the triangles CFA, CFB the side FA is equal to the side FB by construction, the side FC is common, and the two angles at F are right angles. These two triangles, then, have two sides and the angle contained by them equal; they are therefore (Prop. 3.) identical. Consequently the sid CB is equal to the side CA.

Fig. 265.

For the same reason, the triangles CGB, CGD are also identical. Wherefore the sid CD is equal to the side CB, and consequently equal to CA.

And since the right lines CB, CD are equal to the right line CA, it is manifest (Prop. 1. that the circumference which passes through the point A must also pass through th Doint D.

910. PROP. XXXIV. If a radius bisect a chord, it is perpendicular to that chord. If the radius CF (fig. 266.) bisect the chord AB, the angles

DA, CDB are right angles. Draw the radii CA, CB.

In the triangles CDA, CDB the sides CA, CB, being radii, are equal (Prop. 1.), the sides AD, DB are equal by supposition, and the side CD is common. These two triangles, having the three sides equal, are therefore (Prop. 5.) identical. Wherefore the angles CDA, CDB are equal, and consequently (Prop. 10.) are right angles.

COROLLARY. The two angles at C are also (Prop. 5.) equal. Hence it appears, that any angle ACB may be bisected by describing from its vertex C as the centre with any radius AC an are AFB; bisect

Fig. 266.

B

ing the chord of that are AB; and then drawing from the point of division D the right lin CD; for it may then be shown, as in the proposition, that the triangles ACD, DCB an identical, and consequently the angles at C equal.

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