XLVI.

The sine of any arch is equal to half the chord of twice that arch.

For (in the last scheme) AD is the sine of the arch AF, and AF is equal to half the arch AFB and AD half the chord AB, whence the proposition is manifest.

XLVII.

B

D

If two equal and parallel lines AB, CD, be joined by A two others, AC, BD, these will be also equal and parallel. To demonstrate this, join the two opposite angles A and D with the line AD; then it is evident that the line AD divides the quadrilateral ACDB into two triangles ABD, ACD, in which AB is equal to CD byC supposition, and AD is common to both triangles; and since AB is parallel to CD, the angle BAD is equal to the angle ADC (by art. 30), therefore in the two triangles, the sides AB, AD, and the angle BAD are equal respectively to the sides CD, AD, and the angle ADC; hence (by art. 37) BD is equal to AC, and the angle DAC equal to the angle ADB; therefore (by art. 34) the lines BD, AC, must be parallel.

Cor. Hence it follows that the quadrilateral ABDC is a parallelogram, since the opposite sides are parallel. It is also evident that in any parallelogram, the line joining the opposite angles (called the diagonal) as AD, divides the figure into two equal parts, since it has been proved that the triangles ABD, ACD, are equal to each other.

XLVIII.

It follows also from the preceding article, that a triangle ACD (see the preceding figure) on the same base, and between the same parallels with a parallelogram ABDC, is the half of that parallelogram.

XLIX.

From the same article it also follows, that the opposite sides of a paral·lelogram are equal. For it has been proved, that ABDC being a parallelogram, AB is equal to CD, and AC equal to BD.

L.

All parallelograms on the same or equal bases, and between the same parallels, are equal to each other; that is, if BD and GH be equal, and the lines BH, AF be parallel, the parallelograms ABDC, BDFE and EFHG will be equal to each other.

A

C

K

D

E F

For AC is equal to EF each being equal to BD (by art. 49) to both add CE and we have AE equal to CF; therefore in the two triangles ABE, CDF; AB is equal to CD, and AE is equal to CF, and the angle BAE is equal to DCF (by art. 31,) therefore B Ꮐ H the two triangles ABE, CDF are equal (by art. 37) and taking the triangle CKE from both, the figure ABKC is equal to the figure KDFE, to both which add the little triangle KBD, and we have the parallelogram ABDC equal to the parallelogram BDFE. In the same way it may be proved that the parallelogram EFHG is equal to the parallelogram BDFE; therefore the three parallelograms ABDC, BDFE, and EFHG are equal to each other.

Cor. Hence it follows, that triangles on the same base and between the same parallels are equal, since they are the half of the parallelograms on the same base and between the same parallels (by art. 48.)

LI.

In any right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the two sides. Thus if BAC be a right angled triangle the square of the hypotenuse BC, viz. BCMH, is equal to the sum of the squares made on the two sides AB and AC, viz. to ABDE and ACGF.

To demonstrate this, through the point A draw AKL perpendicular to the hypotenuse BC. Join AH, AM, DC, and BG; then it is evident, that DB

D

E

B

F

is equal to BA (by art. 18) and BH equal to BC, therefore in the triangles DBC, ABH, the two legs DB, BC of the one are equal to the two legs AB, BH, of the other; and the included angles DBC and ABH are also equal, (for DBA is equal to CBH being both right, to each add ABC and we have DBC equal to ABH) therefore the triangles DBC, ABH are equal (by art. 37) but the triangle DBC is half of the square ABDE (by art. 48) and the triangle ABH is half the parallelogram BKLH (by the same art.) consequently the square ABDE is equal to the parallelogram BKLH. In the same way it may be proved that the square ACGF is equal to the parallelogram KCML. Therefore the sum of the squares ABDE and ACGF is equal to the sum of the parallelograms BKLH and KCML; but the sum of these parallelograms is equal to the square BCMH, therefore the sum of the squares on AB and AC is equal to the square on BC.

K

H

M

L

Cor. Hence in any right angled triangle, if we have the hypotenuse and one of the legs, we may easily find the other leg, by taking the square of the given leg from the square of the hypotenuse, the square root of the remainder will be the sought leg. Thus if the hypotenuse was 13, and one leg was 5, the other leg would be 12, for the square of 5 is 25, and the square of 13 is 169, subtracting 25 from 169 leaves 144, the square root of which is 12. If both legs are given, the hypotenuse may also be found by extracting the square root of the sum of the squares of the legs; thus if one leg was 6, and the other 8, the square of the first is 36, the square of the second is 64, adding 36 and 64 together gives 100, whose square root is 10, which is the sought hypotenuse.

LII.

Four quantities are said to be proportional, when the magnitude of the first compared with the second is the same as the magnitude of the third compared with the fourth.

Thus 4, 8, 12 and 24, are proportional, because 4 is half of 8, and 12 is half of 24; and if we take equi-multiples Axa, A×b, of the quantities a and b, and other equi-multiples B×a, Bxb, of the same quantities a and b, the four quantities Axa, A×b, B×a, B×b will be proportional, for Axa compared with Axb is of the same magnitude as a compared with b, and B× a compared with Bb is also of the same magnitude as a compared with b.

In

LIII.

any triangle AGg if a line Ee be drawn parallel to either of the sides as Gg, the side Ag will be to AE, as Ag to Ae, or as Gg to Ee.

To demonstrate this, upon the line

AG take the line AB so that a certain multiple of it may be equal to AE, and another multiple of it may be equal to AG; this may be always done accurately when AE and AG are commensurable; if they are not accurately commensurable, the quantity AB be taken so small that certain multiples of it may differ from AE and AG respectively by

may

N

b C d

quantities less than any assignable. On the line AG, take BC, CD, DE, EF, e f y FG, &c. each equal to AB, and through these points draw the lines Bb, Cc, &c. parallel to Gg, cutting the line Ag in the points b, c, d, e, &c. draw also the lines BM, CL, DK, &c. parallel to Ag, cutting the former parallels in the points N, O, P, &c. and the line Gg in the points M, L, K, &c. Then the triangles ABb, BCN, CDO, &c. are similar and equal to each other: for the Jines Bb, CN are parallel, therefore the angle ABb=BCN (by art. 31) and

by the same article the angle BAb is equal to CBN (because BN is parallel to Ab) and by construction AB=BC, therefore (by art. 38) the triangles ABb and BCN are equal to each other; and in the same manner we may prove that the others CDO, DEP, EFQ, &c. are equal to ABb. Therefore Ab÷BN=CO=DP, &c. and Bb-CÑ=D0=EP, &c. but (by art. 49) BN=BC, CO=cd, DP=de; therefore Ab-be=cd=de, &c. and since (by construction) AB=BC=CD, &c. any line AE is the same multiple of AB as the corresponding line Ae is of Ab; and AG is the same multiple of AB as Ag is of Ab; therefore the lines AG, AE, Ag, Ae, are proportional (by art. 52;) that is, AG is to AE as Ag is to Ae; and in a similar manner we may prove that AG is to AE as Gg is to Ee.

LIV.

If any two triangles, ABC, abc, are similar, or have all the angles of the one, equal to all the angles of the other, each to each respectively, that is, CAB=cab, ACB=acb, ABC=abc; the legs opposite to the equal angles will be proportional, viz. AB: ab::AC:ac; AB:ab::BC: bc; and AC: ac:: BC: bc.

To prove this, set off upon a side AB of the largest triangle AE=ab, and through E draw ED parallel to BC, to meet AC. in D, then since DE, BC are parallel, the angle AED is equal to ABC (by art. 31) and this (by supposition) is equal to the angle abc; also the angle DAE is (by A supposition) equal to cab; there

fore in the triangles ADE, abc, the two angles DAE, AED of the one, are equal to the two angles cab, abe of the other, each to each respectively, and the included side AE is (by construction) equal to the included side ab; therefore (by art. 38) AD is equal to ac, and DE equal to be: but since in the triangle ABC there is drawn DE parallel to BC one of its sides, to meet the other two sides in the points DE; therefore (by the preceding art.) AB: AE :: AC : AD, and AB: AE :: BC: DE, and AC : AD :: BC: DE; if in these three proportions for DE we put its equal bc, for AE put ab, and for AD put ac; they will become AB: ab :: AC : ac, and AB : ab : : BC : bc, and AC: ac :: BC: be.

LV.

The chord, sine, tangent, &c. of any arch in one circle, is to the chord, sine, tangent, &c. of the same arch in another, as the radius of the one is to the ra dius of the other.

Let ABD, abd, be two circles; BD, bd, two arches of these circles, equal to one another, or consisting of the same number of degrees — FD, fd, the tangents; Bd, bd, the chords; BE, be, the sines, &c. of these two arches BD, bd, and CD, cd, the radii of the circles; then CD: ed: FD: fd, and CD: cd::" BD: bd, and CD: ed: BE: be, &c. For since the arches BD, bd, are equal, the angles BCD, bcd, are also equal, and FD, fd, being tangents to the points D and d, the angles CDF, cdf are each equal to a right angle (by art. 22;)

therefore since in the two triangles CDF, cdf, the two angles FCD, CDF of the one, are equal to the two angles fed, cdf, of the other, each to each, the remaining angle CFD is also equal to the remaining angle cfd, (by art. 36 ;) consequently the triangles CFD, cfd, are similar. The triangles BCD, bed are also similar, for the angle CBD is equal to the angle CDB, being each subtended by the radius; therefore (by art. 36) each of these angles is equal to half the supplement of the angle BCD; and in the same manner the angle cbd or cdb is equal to half the supplement of the angle bed, and since the angle BCD is equal to bcd, the angles of these two triangles must be equal, consequently they are similar. The triangles BCE, bce are also similar, because BE is parallel to, FD, and be parallel to fd. Hence we obtain (by art. 54) the following analogies. CD:cd::FD: fd; CD: cd :: BD: bd; CB: cb:: BE: be, &c.

LVI.

A

H

B

Let ABD be a quadrant of a circle, described by the radius CD, BD any arch of it, BA its complement, BG or CF the sine, CG or BF the co-sine, DE the tangent, AH the co-tangent, CE the secant, and CH the co-secant of that arch BD. Then since the triangles CDE, CGB, are simi-F lar or equi-angular we shall have (by art. 54) DE: CE :: BG: CB, that is, the tangent of an arch, is to secant of the same, as the sine of it is to radius. Also, CE: CD::CB: CG; that is, the secant is to radius as the radius to the cosine of an arch. Also, CF: CA:: CB: CH, that is, the C sine is to radius as radius to the co-secant of an arch; and since the triangle CAH is similar to the triangle CDE, we have AH: CA :: CD: DE, that is, the co-tangent is to the radius as the radius to the tangent of an arch.

LVII.

G

D

In all circles, the sine of 90°, the tangent of 450, and the chord of 60°, are each equal to the radius.

D

F

G

E

B

For in the circle DFAEB, let the arch BE be 450, the arch BA 60°, and BF 90°. Draw through the centre C the diameter DCB and perpendicular thereto the tangent BG meeting CE produced in G; draw the chord BA, and join CF, CA.— Then since the arch BF is 90°, DF must be 90°, whence (by art. 12 & 19) the radius CF is equal to the sine of the arch BF, or sine of 90°. Again, in the triangle CBG, since the angle CBG is 90°, and BCG is 45° by supposition, the angle CGB is also 450 (by art. 36) therefore (by art. 39) BG is equal to CB, that is, the tangent of 450 is equal to the radius. Again, the angle ACB is 60° (being measured by the arch BA) and the angle CBA is also 600 (being measured by half the arch AD=120° by art. 40) therefore (by art. 39) CA=AB, that is, the chord of 60° is equal to the radius.

The four following propositions contain the demonstration of the rules by which all the calculations of trigonometry may be made; they were inserted here in order to prevent any embarrassment of the young calculator, from the introduction of the demonstrations among the precepts for calculation.

LVIII. In any plane triangle, the sides are proportional to the sines of the opposite angles.

Let ABC be the triangle; produce the lesser side AB to F, making AF equal to BC; from B and F let fall the perpendiculars BD, FE, upon AC (produced if necessary;) then FE is the sine of the angle A A, and BD is the sine of the angle C, the

B

DE

tadius being BC equal to AF; now the triangles ABD, AFE, having the angle A common to both, and the " angle D equal to the angle E (being each equal to a right angle) are similar; hence (by art. 54) as AF (or its equal BC) is to AB, so is FE to BD; that is, BC is to AB as the sine of the angle A is to the sine of the angle C.

LIX.

E

In any triangle (supposing any side to be the base, and calling the other two the sides) the sum of the sides is to their difference, as the tangent of half the sum of the angles at the base is to the tangent of half the difference of the same angles.

Thus, in the triangle ABC, if we call AB the base, it will be as the sum of AC and CB is to their difference, so is the tangent of half the sum of the angles ABC, BAC, to the tangent of half their difference.

I

D

H

F

B

Dem. With the longest leg CB as radius, describe a circle about the centre C, meeting the shorter side AC (produced on each side) in the points D and E, join EB, DB; draw AH perpendicular to DB, and AF perpendicular to EB; then (by art. 42) the angle EBD, being in a semi-circle, is a right angle; and the triangles AHD, AFE, are similar, and AF is equal to HB. Moreover, since CB is equal to CD or CE, AD is the sum and AE is the difference of the legs AC, CB; likewise (by art. 33) the angle BCD is equal to the sum of the angles BAC, ABC, and therefore (by art. 40) the angle DEB, or its equal DAH, is equal to half the sum of the angles at the base ABC, BAC. Again (by art. 33) the angle BAC is equal to the sum of the angles CEB (or CBE) and ABE, and therefore is equal to the sum of the angle ABC, and twice the angle ABE; hence the angle ABE or its equal BAH, is equal to half the difference of the angles at the base. But in the right angled triangles AHD, AHB, making AH radius, the legs DH, HB are the tangents of the angles DAH, BAH, or the tangents of half the sum and half the difference of the angles at the base; but by reason of the similar triangles AHD, AFE, we have AD AE :: DH: AF or HB; that is, AD, the sum of the legs AC and CB, is to AE their difference, as DH the tangent of half the sum of the angles at the base (the radius being AH) is to HB the tangent of half the difference of the same angles, (to the same radius,) and therefore (by art. 55) as the tabular tangent of half the sum of the angles at the base is to the tabular tangent of half the difference of the same angles.