! A : B C F G F G M B D But the angle æ is equal to the alternate angle B, and the angle y to the alternate angle C. Therefore, substituting B for x, and C for y, the three angles B, A, C are together equal to two right angles. COROLLARY. Hence, if two angles of any triangle be known, the third is also found; since the third angle is that which the other two taken together want of two right angles. 898. Prop. XXIII. If two triangles have two angles equal, they have also the third angle 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 must be equal to the angle G. Fig. 256. 899. Prop. XXIV. The exterior angle of any triangle is equal to the two interior and 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. Fig. 257. 900. Prop. XXV. Triangles which have two angles and the side which lies between them equal are identical. In the two triangles BAC, FDG (fig. 258.), if the angle F is equal to the angle B, the 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. 901. Prop. XXVI. If two angles of a triangle are equal, the sides opposite to those angles are also equal. Conceive the angle A (fig. 259.) to be bisected by the line AD. 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 cominon to the two triangles. These two triangles, therefore, having two angles and the side Fig. 259. which lies between them equal, are (Prop. 25.) identical. Wherefore the side AB is equal to the side AC. 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. Also, the side BD is common to the two triangles BAD, BCD. These two triangles have then two angles and the side which lies between them equal, and are therefore (Prop. 3.) identical. Wherefore the side AB opposite to the angle n is (Prop. 26.) equal to the side DC opposite to the angle m; and the side BC opposite to the angle s is equal to the side AD opposite to the equal Angle r. B с Fig. 258, A B D B A Fig. 260. B с L M A D B с F L A D COROLLARY. Hence, the diagonal bisects the parallelogram; for the triangles BAD, BCD, having the three sides equal, are identical. 903. Prop. XXVIII. Parallelograms which are between the same parallels, and have the same base, are equal. 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 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 between the same parallels, the triangle is equal to half the parallelogram. 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 paral. lelogram AFGD. But the parallelogram AFGD is equal to the parallelogram ABCD, because these two parallelograms are upon the same base, and between the same parallels; therefore the triangle AFD is equal to half the parallelogram ABCD. 905. Prop. XXX. Parallelograms which are between the same parallels, and have equal bases, are equal. Let the two parallelograms ABCD, LFGM (fig. 263.) be between the same parallels BG, AM, and have the equal bases AD, LM; these two parallelograms are equal. Draw the lines AF, DG. 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 Fig. 263, 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 parallelogram. Now the paralle am 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 two parallelograms are between the same parallels, and have the same base FG. The parallelogram ABCD is therefore equal to the parallelogram LFGM. 906. 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. The two parallelograms ABCD, LFGM are equal (Prop. 30.), because they are between the same 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; these two triangles are therefore equal. 907. Prop. XXXII. In a right-angled triangle, the square of the hypotenuse, or side subtending 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. в с F G A D L M M N P с 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 hali 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. 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 are 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 angle x, therefore these two angles, contained between the equal sides MB, BC and AB, BÒ, 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 the rectangle BDGP; the square and the rectangle are therefore equal. In the same manner it may be demonstrated that the square ARNC and the rectangle CFGP are equal. Wherefore it follows that the whole square BDFC is equal to the two squares MLAB, ARNC taken together. G F CIRCLES. ; B 908. DEFINITIONS. — 1. A right line (fig. Prop. 39. AB) terminated both ways by the circumference of a circle is called a chord. 2. A line (fig. Prop. 39. AB) which meets the circumference in one point only is called 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 circle 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 segment 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 A 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 Fig. 265. contained by them equal; they are therefore (Prop. 3.) identical. Consequently the side CB is equal to the side CA. For the same reason, the triangles CGB, CGD are also identical. Wherefore the side 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 the point 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 CDA, 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 A Can arc AFB; bisecting the chord of that arc AB; and then drawing from the point of division D the right line CD; for it may then be shown, as in the proposition, that the triangles ACD, DCB are identical, and consequently the angles at C equal. D B Fig. 266. G с B D FM BD Fig. 267. Fig. 268. D 911. Pror. 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 divi. sion 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. 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. Let ABDF be the arc 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 arc. 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. 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 Fig. 269. three sides equal ; they are therefore (Prop. 5.) identical. Wherefore the two angles at F are 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 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 jg equal to the right line DB, and the arc AF equal to the arc FB. Draw the right lines CA, CB. 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 Dare 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.) Fig. 270. 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 arc AFB are therefore bisected by the radius CF. 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. 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 there. fore a tangent to the circle. COROLLARY. It follows, therefore, that a perpendicular is the shortest line that can be centre. с D B F D B Fig. 271. T B Fig. 27%. a B T Fig. 273. B T A drawn from any point to a given line ; since the perpendicular CT is shorter than any other line which can be drawn from the point C to the line AB. 916. Prop. XL. If a right line be drawn touching a circumference, a radius drawn to the point of contact will be perpendicular to the tangent. Let the line AB (fig. 272.) touch the circumference of a circle A 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 A B, 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. 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 by half the arc 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 arc TFD. TF is therefore half the arc TFD. 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. But the angle C is 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 TGD, 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 arc CFD by which it is subtended. Suppose a tangent passing through the point T. The three angles at T are measured by half the circumference (Prop. 22.), but the angle ATD is measured (Prop. 41.) by half Fig. 274. 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 of 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 arc AB, the angle ACB is double of the angle ADB. For the angle ACB is measured by the arc AB, and the angle ADB is (Prop. 42.) measured by half the same arc AB; the angle ACB is therefore double of the angle ADB. 920. Prop. XLIV. Upon a given line, to describe a segment of a circle containing a given angle. Let AB (fig. 276.) be the given line and G the given angle, it is required to draw surb a circumference of a circle through the points A and B that the angle D shall be equal to 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. 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 Fig. 276. LAB. But the angle LAB is made equal to the angle G; the angle D is therefore equal to the angle G. 921. Phor. XLV. In every triangle the greater side is opposite to the greater angle, and the greater angle to the greater side. In the triangle ABC (fig. 277.), if the side A B be greater than the side AC, the angle F 1 Pig. 275. B |