PROP. III The sum of the sixes of any two arches of a circle, is to the difference of their sines, as the tangent of half the sum of the arches to the tangent of half their difference. Let AB, AC be two arches of a circle ABCD; let E be the centre, and AEG the diameter which passes through A: sin. AC+ sin. AB: sin. AC-sin. AB :: tan. (AC+AB): tan. (AC—AB). Draw BF parallel to AG, meeting the circle again in F. Draw BH and CL perpendicular to AE, and they will be the sines of the arches AB and AC; produce CL till it meet the circle again in D; join DF, FC, DE, EB, EC, DB. F B K E LH Now, since EL from the centre is perpendicular to CD, it bisects the line CD in Land the arch CAD in A DL is therefore equal to LC, or to the sine of the arch AC; and BH or LK being the sine of AB, DK is the sum of the sines of the arches AC and AB, and CK is the difference of their sines; DAB also is the sum of the arches AC and AB, because AD is equal to AC, and BC is their difference. Now, in the triangle DFC, because FK is perpendicular to DC, (3. cor. 1.) DK: KC = :: tan. DFK : tan. CFK; but tan. DFK=tan. arc. BD, because the angle DFK (20. 3.) is the half of DEB, and is therefore measured by half the arch DB. For the same reason, tan. CFK tan. arc. BC; and consequently, DK: KC :: tan. arc. BD: tan. arc. BC. But DK is the sum of the sines of the arches AB and AC; and KC is the difference of the sines; also BD is the sum of the arches AB and AC, and BC the difference of those arches. Therefore, &c. Q. E. D. COR. 1. Because EL is the cosine of AC, and EH of AB, FK is the sum of these cosines, and KB their difference; for FK=1 FB+ EL - EH+EL, and KB= LH = EH-EL. Now, FK: KB :: tan. FDK: tan. BDK; and tan. FDK=cotan. DFK, because DFK is the complement of FDK; therefore, FK: KB:: cotan. DFK : tan. BDK, that is, FK: KB :: cotan. arc. DB: tan. arc. BC. The sum of the cosines of two arches is therefore to the difference of the same cosines as the cotangent of half the sum of the arches to the tangent of half their difference. COR. 2. In the right angled triangle FKD, FK: KD R tan. DFK : Now FK=cos. AB + cos. AC, KD=sin. AB+sin., AC, and tan. DFK = tan. (AB + AC), therefore cos. AB + cos. AC : sid. AB+sin. AC :: R: tan. (ÁB + AC). In the same manner, by help of the triangle FKC, it may be shown that cos. AB+cos. AC: sin. AC-sin, AB :: R: tan. AC-AB), COR. 3. If the two arches AB and AC be together equal to 90°, the tangent of half their sum, that is, of 45o, is equal to the radius, And the arch BC being the excess of DC above DB, or above 90°, the half of the arch BC will be equal to the excess of the half of DC above the half of DB, that is, to the excess of AC above 45°; therefore, when the sum of two arches is 90°, the sum of the sines of those arches is to their difference as the radius to the tangent of the difference between either of them and 45°. PROP. IV. The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference. Let ABC be any plane triangle ; CA+AB CA-AB: tan. (B+C): tan. (B-C). For (2.) CA: AB :: sin. B: sin. C and therefore (E. 5.) ; CA+AB: CA-AB :: sin. B+sin. C: sin. B-sia. C. : : tan. (B+C) tan. (B-C); therefore also, (11. 5.) CA+AB: CA-AB:: tan (B+C): tan. (B-C). Q. E. D. Let ABC be a triangle; the sum of AB and AC any two sides, is to the difference of AB and AC as the tangent of half the sum of the angles ACB and ABC, to the tangent of half their difference. About the centre A with the radius AB, the greater of the two sides, describe a circle meeting BC produced in D, and AC produced in E and F. Join DA, EB, FB; and draw FG parallel to CB, meeting EB in G. Because the exterior angle EAB is equal to the two interior ABC, ACB, (32. 1.): and the angle EFB, at the circumference is equal to half the angle EAB at the centre (20. 3.); therefore EFB is half the sum of the angles opposite to the sides AB and AC. Again, the exterior angle ACB is equal to the two interior CAD, ADC, and therefore CAD is the difference of the angles ACB, ADC, that is of ACB, ABC, for ABC is equal to ADC. Wherefore also DBF, which is the half of CAD, or BFG, which is equal to DBF, is half the difference of the angles opposite to the sides AB, AC. Now because the angle FBE in a semicircle is a right angle, BE is the tangent of the angle EFB, and BG the tangent of the angle BFG to the radius FB ; and BE is therefore to BG as the tangent of half the sum of the angles ACB, ABC to the tangent of half their difference. Also CE is the sum of the sides of the triangle ABC, and CF their difference; and because BC is parallel to FG, CE: CF :: BE: BG, (2. 6.) that is, the sum of the two sides of the triangle ABC is to their difference as the tangent of half the sum of the angles opposite to those sides to the tangent of half their difference. Q. E. D. PROP. V. THEOR. If a perpendicular be drawn from any angle of a triangle to the opposite side, or base; the sum of the segments of the base is to the sum of the other two sides of the triangle as the difference of those sides to the difference of the segments of the base. For (K. 6.), the rectangle under the sum and difference of the segments of the base is equal to the rectangle under the sum and differ ́ence of the sides, and therefore (16. 6.) the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base. Q. E. D. PROP. VI. THEOR In any triangle, twice the rectangle contained by any two sides is to the difference between the sum of the squares of those sides, and the square of the base, as the radius to the cosine of the angle included by the two sides. Let ABC be any triangle, 2AB.BC is to the difference between AB2+BC2 and AC2 as radius to cos. B. From A draw AD perpendicular to BC, and (12. and 13. 2.) the difference between the sum of the squares of AB and BC, and the square on AC is equal to 2BC.BD. But BC.BA: BC.BD:: BA: BD :: R cos. B, therefore also 2BC.BA: B A D 2BC.BD: R: cos. B. Now 2BC.BD is the difference between AB2 +BC2 and AC2, therefore twice the rectangle AB. BC is to the difference between AB2+BC2, and AC2 as radius to the cosine of B. Wherefore, &c. Q. E. D. B A COR. If the radius=1, BD=BA Xcos. B, (.), and 2BC.BAX cos. B=2BC.BD, and therefore when B is acute, 2BC.BAXcos. B=BC2 +BA2 - AC2, and adding AC2 to both; AC2+2 cos. B×BC.BA BC+BA; and taking 2. cos. BxBC.BA from both, AC2=BC2-2 cos. BXBC.BA+BA2. Wherefore AC✓ (BC2-2 cos. BXBC. BA+BA2). D If B is an obtuse angle, it is shewn in the same way that AC= ✔ (BC2+2 cos. BBC.BA+BA2). PROP. VII. Four times the rectangle contained by any two sides of a triangle, is to the rectangle contained by two straight lines, of which one is the base or third side of the triangle increased by the difference of the two sides, and the other the base diminished by the difference of the same sides, as the square of the radius to the square of the sine of half the angle included between the two sides of the triangle. Let ABC be a triangle, of which BC is the base, and AB the greater of the two sides; 4AB.AC : (BC+(AB~AC))×(BC—(AB—AC)): (sin. BAC). Produce the side AC to D, so that AD=AB; join BD, and draw AE, CF at right angles to it; from the centre C with the radius CD describe the semicircle GDH, cutting BD in K, BC in G, and meeting BC produced in H. It is plain that CD is the difference of the sides, and therefore that BH is the base increased, and BG the base diminished by the difference of the sides; it is also evident, because the triangle BAD is isosceles, that DE is the half of BD, and DF is the half of DK, wherefore DE-DF the half of BD-DK, (6.5.) that is EF-1 BK. And because AE is drawn parallel to CF, a side of the triangle CFD, AC: AD :: EF: ED, (2. 6.); and rectangles of the same altitude being as their bases AC.AD: AD2 :: EF.ED : ED, (1. 6.), and therefore 4AC.AD : AD2 :: 4EF.ED: ED2, or alternately, 4AC.AD: 4EF.ED :: AD2 : ED". But since 4EF=2BK, 4EF.ED=2BK.ED=2ED.BK=DB.BK= HB.BG; therefore, 4AC.AD: DB.BK :: AD2 : ED3. Now AD : ED: R sin. EAC=sin. BAC (1. Trig.) and AD2 : ED2 :: R2 : (sin.BAC): therefore, (11. 5.) 4AC.AD: HB.BG :: R2 : (sin. BAC), or since AB-AD, 4AC.AB: HB.BG :: R2: (sin 1 BAC)o. Now 4AC.AB is four times the rectangle contained by the sides of the triangle; HB.BG is that contained by BC+(AB-AC) and BC(AB-AC). Therefore &c. Q. E. D. Four times the rectangle contained by any two sides of a triangle, is to the rectangle contained by two straight lines, of which one is the sum of those sides increased by the base of the triangle, and the other the sum of the same sides diminished by the base, as the square of the radius to the square of the cosine of half the angle included between the two sides of the triangle. Let ABC be a triangle, of which BC is the base, and AB the greater of the other two sides, 4AB.AC: (AB+ÁC+BC) (AB+AC−BC) :: R2 (cos. BAC)2. From the centre C, with the radius CB, describe the circle BLM, meeting AC, produced, in L and M. Produce AL to N, so that AN. =AB; let AD AB; draw AE perpendicular to BD; join BN, and |