C opposite 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, A C, B. that the arc ADB is greater than the arc AFC; and consequently D the angle at the circumference C, which is measured (Prop. 42.) s 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 \l. ~/ opposite to the angle C will be greater than the side AC opposite B -to the angle B. Fig.277. The angle C is measured (Prop. 42.) by half the arc ADB, and the angle B by half the arc AFC. But the angle C is greater than the angle B; the arc ADB is therefore greater than the arc AFC, and consequently the chord AB is greater than the chord A.C. 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 B.C. 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 arc AC; and the angle at the circumference BCD is measured by half the arc BD; the arcs AC, BD are therefore equal. 923. Pror. XLVII. If a tangent and chord be parallel to each other, they intercept equal arcs. Let the tangent FG (fig. 279.) be parallel Fig.278. Fig.279. to the chord AB; the arc TA will be equal to the arc TB. 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 are TB. The halves of the arcs TA, TB, and consequently the arcs themselves, are therefore equal. 924. Paor. 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, D F (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 G p 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 circumference GAB is measured (Prop. 42.) by half the arc GFB. The angle FCB is therefore also measured by half the arc GFB. Because the chords AG, DF are parallel, the arcs GF, AD are loo – ~ (Prop. 46.) equal: AD may therefore be substituted in the room Fig. 280. of GF; wherefore the angle FCB is measured by half the arcs AD, FB. 925. Pror. 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 F.B. Because the chords GC, D F are parallel, the arcs GD, CF are (Prop. 46.) equal; the arc FB is therefore the difference of the arc G D and the arc CF B. Where the angle A has for its measure half the difference of the arcs GD, CFB. 926. Pror. L. The angle formed by two tangents is measured by half the difference of th’ tuco 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 A.C. 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 D.F, is measured (Prop. 41.) by half the arc F.D. Therefore the angle CAB is also measured by half the arc F.D. Because the tangent AC and the chord DF are parallel, the intercepted arcs GF G D are (Prop. 47.) equal. The arc FD is there. fore the difference between the arc GLD and the arc GFD. Therefore the angle CAB, which is measured by half the arc FI), is also A. measured by half the difference of I) the arcs GLD, GFD. Fig. 282. Fig. 283. CoRoll. ARY. In the same way it may be demonstrated that the angle formed by a tangent ATC (fig. 283.) and a secant ADB is measured by half the difference of the two intercepted arcs. 927. PRor. LI. To raise a perpendicular at the ertremity of a giren line. At the extremity A (fig. 284.) of the given line A B let it be required to raise a per pendicular. D -From any point C taken above the line AB describe a circum- * F ference passing through the point A and cutting the line A B in any \c other point, as G. Draw the diameter DG and the right line AD; \ \| \, . this line AD will be perpendicular to the line A B. The angle DAG at the circumference is measured (Prop. 42.) by half the arc DFG, which is half the circumference, because DCG is A== a diameter. The angle DAG is therefore measured by one fourth Fig. 284. part of the circumference, and consequently (Defin. 10.) is a right angle, whence the line AD is (Prop. 11.) perpendicular to the line A B. Conoll ARY. Hence it follows that the angle at the circumference which is subtended by a diameter must be a right angle. 928. Pitor. 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. 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 perpendicular to the extremity of the radius CT, and consequently (Prop. 40.) is a tangent to the circle DTB. suftraces. 929. Definitions.—1. A mathematical point has neither length, breadth, nor thickness. The physical point, now for consideration, has a supposed length and breadth exceedingly small. 2. A physical line is a series of physical points, and consequently its breadth is equal to that of the physical points whereof it is composed. 3. Since physical lines are composed of points, as numbers are composed of units, points may be called the units of lines. 4. As to multiply one number by another is to take or repeat the first number as many times as there are units in the second; so to multiply one line by another is to take or repeat the first line as many times as there are units, that is, physical points, in the second. 930. Poor. 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. If as many physical lines equal to AB as there are physical points in the line AD be raised perpendicularly — upon AI), these lines AB, al., &c. will fill up the whole * * * * * * * * surface of the rectangle ABCD. Wherefore the surface Fig. 286. ABCD is equal to the line AB taken as many times as there are physical points in the line AD; that is, (Defin. 4.) equal to the line A B multiplied by the line AD. 931. Pror. LIV. The surface of a triangle is equal to half the product of its altitude and base. If from the vertex of any angle A (fig. 287.) of the triangle BAC be drawn AD, Perpendicular 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 - equal to half the product of the height AD and the base BC. -------. ... 6. 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 CC. 932. Pror. LV. To measure the surface of any rectilineal figure. 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 ACF, FCD be found. These three surfaces added together will give the whole surface of the figure ABCDFA. 933. Prop. 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 Fig. 288. its circumference, the half of the product will give the surface of the circle. 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 Fig. 287. 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 N 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 c ) and its base. ! To find the surface of all the small triangles whereof the circle is com- / posed, multiply the height, that is, the radius, by all the bases, that is, by `i -the circumference, and take the half of the product; whence the area or Fig.289. surface of the circle will be equal to half the product of the radius and circumference. 934. PRor. LVII. To draw a triangle equal to a given circle. Let it be required to form a triangle the surface of which shall be equal to that of the At the extremity of any ra dius CA of the circle, raise a face of the triangle BCA will be equal to that of the circle AG FIDA. Fig. 290. 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. DeriNitions. – 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. Phor. LVIII. Parallelograms which are between the same parallels are to one anLet the two parallelograms ABCD, FGLM (fig. 291.) be between the same parallels other as their bases. BL, AM, the surface of the parallelogram ABCD contains P & co. o: i. the surface of the parallelogram FGLM as many times exactly as the base AD contains the base FM. Suppose, for example, that the base AD is triple of the base FM ; in this case the surface ABCD will also be triple of the surface FGLM. Divide the base AD into three parts, each of which is equal to the base FM, and draw from the points of divi- Fig. 291. sion the lines NP, RS parallel to the side A B. The parallelograms ABPN, FGLM being between the same parallels and having equal bases, the parallelogram ABPN is (Prop. 30.) equal to the parallelogram FGLM. For the same reason, the parallelograms NPSR, RSCD are also equal to the parallelogram FGLM. The parallelogram ABCD is therefore composed of three parallelograms, each of which is equal to the parallelogram FGLM. Consequently the parallelogram ABCD is triple of the parallelogram FGLM. 937. Pror. LIX. Triangles which are between the same parallels are to one another as their bases, Let the two triangles ABC, DFG (fig. 292.) be between the same parallels LF, AG, the surface of the triangle ABC contains the surface of the 1. M. F. triangle DFG as many times as the base AC contains the —k base DG. Suppose, for example, that the base AC is triple of the base DG, in this case the surface ABC will be triple of the surface DFG. \ Divide the base AC into three equal parts, AN, N R, - \ RC, each of which is equal to the base DG, and draw the N r right lines BN, B.R. Fig. 292. The triangles ABN, DFG being between the same parallels and having equal bases, the triangle ABN is (Prop. 31.) equal to the triangle DFG. For the same reason, the triangles NBR, RBC are each equal to the triangle DFG. The triangle ABC is therefore composed of three triangles, each of which is equal to the triangle DFG. Wherefore the triangle ABC is triple of the triangle DFG. 938. Pror. LX. If a line be drawn in a triangle parallel to one of its sides, it will cut the other two sides proportionally. In the triangle BAC (fig. 293.), if the line DF be parallel to the side BC, it will cut the other two sides in such manner that the segment AD will be to the A segment DB as the segment A F is to the segment FC. Suppose, for instance, the segment AD to be triple of the segment DB, the segment AF will be triple of the segment FC. Draw the diagonals DC, F.B. The triangles AFD, DFB are between the same parallels, as will r be easily conceived by supposing a line drawn through the point F parallel to the side AB. These two triangles are therefore to one so another (Prop. 59.) as their bases; and since the base AD is triple of the base DB, the triangle AFD will be triple of the triangle DFB. Again, the triangles BFD, FDC are between the same parallels DF, BC, and upon the same base DF. These two triangles are therefore (Prop. 31.) equal; and since the triangle AFD is triple of the triangle DFB, it will also be triple of the triangle FDC. Lastly, the triangles ADF, FDC are between the same parallels, as will be easily conceived by supposing a line drawn through the point D parallel to the side AC. These two triangles are therefore to one another (Prop. 59.) as their bases; and since the triangle ADF is triple of the triangle FDC, the base AF will be triple of the base FC. 939. Prop. LXI. Equiangular triangles have their homologous sides proportional. In the two triangles ABC, CDF (fig. 294.), if the angle A be G equal to the angle C, the angle B equal to the angle D, and the angle C equal to the angle F: the side AC, for example, opposite to the angle B is to the side CF opposite to the angle D as the side AB opposite to the angle C is to the side CD opposite to the angle F. Place the two triangles so that the sides AC, CF shall form one right line, and produce the sides AB, FD till they meet in G. The interior and exterior angles GAF, DCF being equal, the A C r lines GA, DC are (Prop. 20.) parallel. In like manner, the alter- Fig.294. mate angles GFA, BCA on the same sides being equal, the lines GF, BC are (Prop. 20.) parallel. Wherefore the quadrilateral figure BG DC is a parallelogram, and consequently its opposite sides are equal. In the triangle GAF the line BC, being parallel to the side FG, cuts (Prop. 60.) the other two sides proportionally; that is, AC is to CP as AB is to BG, or its equal CD. 940. Prior. LXII. Triangles the sides of which are proportional are equiangular. In the two triangles BAC, FDG (fig. 295.), if the A side AB is to the side DF as the side BC is to the side FG and as the side AC to the side DG, these two triangles have their angles equal. Let the side A B be supposed triple of the side DF; the side AC must be triple of the side DG, and the side BC triple of the side FG. If the triangle FDG be not equiangular with the tri- angle BAC, another triangle may be formed equiangular B C F G with it; for example, FL.G. But this is impossible; Fig. 295. for if the two triangles BAC, FLG were equiangular, their sides would be (Prop. 61.) proportional; and BC being triple of FG, A B would be triple of L.F. But AB is triple of DF; whence L.F would be equal to DF. For the same reason, LG would be equal to DG. Thus, the two triangles FLG, FDG, having their three sides equal, would be (Prop. 5.) identical; which is absurd, since their angles are unequal. 941. Paor. LXIII. Triangles which have an angle in one equal to an angle in the other, and the sides about these angles proportional, are equiangular. If in the two triangles BAC, NMP (fig. 296.) the angle A be equal to the angle M, and the side AB be to the side MN as the side AC is to the side MP, the two triangles are equiangular. A. If AB be triple of MN, AC must be triple of MP. Now, if the angle MNP, for example, is not equal to the angle ABC, another angle may be made, as MNIR, which shall be equal to it. But this is impossible; for the two triangles BAC, NMR, having two angles equal, would be equiangular, and consequently (Prop. 61.) would have their sides proportional ; wherefore, AB b * {= being triple of MN, AC would be triple of MR, which Fig. 296. cannot be, since AC is triple of MP. 942. Pror. LXIV. A right line which bisects any angle of a triangle dirides the side opposite to the bisected angle into two segments, which are proportional to the two other sides. In the triangle BAC, let the angle BAC be bisected by the right line AD, making the angle r equal to the angle s. The segment BD is to the segment DC as the side BA to the side A.C. / Produce the side BA, and draw CF parallel to DA. The lines DA, CF being parallel, the interior and exterior angles r, F are (Prop. 19.) equal, and the alternate angles s, C are (Prop. 17.) A also equal. And since the angle r is equal to the angle s, the angle F a/ ! will also be equal to the angle C ; and consequently the side AF is s \ equal to the side A.C. /* In the triangle BFC, the line AD being parallel to the side FC; > * BD (Prop. 60.) will be to DC as BA is to AF, or its equal AC. B d +. 943. Pror. LXV. To find a fourth proportional to three given lines. Fix. 297. Let the three lines be A, B, C (fig. 298.), it is required to find a fourth line D, such that the line A shall be to the line B as the line C is to A– the line D. b BForm any angle RFG, make FM equal to the line C– _2^ A, MG equal to the line B, and FN equal to the line D-------- L C; draw the right line MN, and through the point G __ draw GL parallel to MN; NL will be the fourth pro- portional required. --> N In the triangle FLG the line NM, being parallel to f M G the side LG, cuts the other two sides (Prop. 60.) propor- Fig.298. tionally. Wherefore FM is to MIG as FN is to NL; that is, A is to B as C is to D, 944. Paor, LXVI. To find a third proportional to two given lines. Let the two lines be A, B (fig. 299.), it is required to A find a third line C, such that the line A shall be to the B line B as the line B is to the line C. C-------- In the triangle FLG the line NM, being parallel to the side LG, cuts the other two Y |