Let AC and CB be two unequal magnitudes, of which AC is the greater, and AB the sum. Bisect AB in D; and to AD, DB, which are equal, let DC be added; then AC will be equal to BD and DC together; that is, to BC and twice. DC; consequently twice DC is the difference, and DC half that difference; but AC the greater is equal to AD, DC; that is, to half the sum added to half the difference, and BC the less is equal to the excess of BD, half the sum, above DC half the difference. Therefore, &c. Q. E. D. COR. Hence, if the sum and difference of two magnitudes be given, the magnitudes themselves may be found; for to half the sum add half the difference, and it will give the greater; from half the sum subtract half the difference, and it will give the less. PROP. III. FIG. 8. In a plane triangle, the sum of any two sides is to their difference, as the tangent of half the sum of the angles at the base, to the tangent of half their diffe rence. Let ABC be a plane triangle, the sum of any two sides AB, AC will be to their difference as the tangent of half the sum of the angles at the base ABC, ACB, to the tangent of half their difference. About A as a centre, with AB the greater side for a distance, let a circle be described, meeting AC produced in E, F, and BC in D; join DA, EB, FB: and draw FG parallel to BC, meeting EB in G. The angle EAB (32. 1.) is equal to the sum of the angles at the base, and the angle EFB at the circumference is equal to the half of EAB at the centre, (20. 3.); therefore EFB is half the sum of the angles at the base; but the angle ACB (32. 1.) is equal to the angles CAD and ADC, or ABC together; therefore FAD is the difference of the angles at the base, and FBD at the circumference, or BFG, on account of the parallels FG, BD, is the half of that difference; but since the angle EBF in a semicircle is a right angle, (def. 7.) FB being radius, BE, BG are the tangents of the angles EFB, BFG; but it is manifest that EC is the sum of the sides BA, AC, and CF their difference; and since BC, FG are parallel, (2. 6.) EC is to CF, as EB to BG; that is, the sum of the sides is to their difference, as the tangent of half the sum of the angles at the base to the tangent of half their difference. 467 * PROP. IV. FIG. 8. In a plane triangle, the cosine of half the difference of any two angles is to the cosine of half their sum, as the sum of the opposite sides to the third side; and the sine of half the difference of any two angles is to the sine of half their sum, as the difference of the opposite sides to the third side. Let ABC be a plane triangle, then, cos (CB) : cos! (C+B):: BA+AC: BC, and sin (CB): sin (C+B) :: BA AC: BC. For, in the preceding proposition, it was shown, that EFB is equal to (C+B), and that CBF is equal to (CB); and since EBF is a right angle, CBE is the complement of CBF, and E the complement of BFE. Now, in the triangle CBE, sin CBE sin E:: CE: BC; that is, cos (CB): cos (C+B): AB+ AC: BC. : Again, in the triangle CBF, sin CBF: sin CFB:: CF: BC; that is, sin (CB): sin (C+B) : : AB-AC: BC. Therefore, &c. Q. E. D. PROP. V. FIG. 18. In any plane triangle BAC, whose two sides are BA, AC, and base BC, the less of the two sides, which let be BA, is to the greater AC as the radius is to the tangent of an angle, and the radius is to the tangent of the excess of this angle above half a right angle as the tangent of half the sum of the angles B and C at the base, is to the tangent of half their difference. At the point A, draw the straight line EAD perpendicular to BA; make AE, AF, each equal to AB, and AD to AC; join BE, BF, BD, and from D, draw DG perpendicular to BF. And because BA is at right angles to EF, and EA, AB, AF are equal, each of the angles EBA, ABF is half a right angle, and the whole EBF is a right angle; also (4. 1. El.) EB is equal to BF. And since EBF, FGD are right angles, EB is parallel to GD, and the triangles EBF, FGD are similar; therefore EB is to BF, as DG to GF, and EB being equal to BF, FG must be equal to GD. And because BAD is a right angle, BA the less side is to AD or AC the greater as the radius is to the tangent of the angle ABD; and because BGD is a right angle, BG is to GD or GF as the radius is to the tangent of GBD, which is the excess of the angle ABD above ABF half a right angle. But because EB is parallelto GD, BG is to GF as ED is to DF, (2. 6.) that is, (since ED is the sum of the sides BA, AC, and FD their difference, (3. of this,)) as the tangent of half the sum of the angles B, C, at the base to the tangent of half their difference. Therefore, in any plane triangle, &c. Q. E. D. PROP. VI. FIG. 9. & 10. In any triangle, twice the rectangle contained by any two adjacent sides is to the difference between the sum of the squares of these two sides, and the square of the base, as the radius is to the cosine of the angle included by the two sides. Let ABC be a plane triangle, twice the rectangle contained by AB, BC, is to the difference between the sum of the squares of AB, BC, and the square of the base AC, as the radius to the cosine of the angle ABC. From A draw AD perpendicular to BC, then (by 12. and 13. 2. El.) the difference between the sum of the squares of AB, BC, and the square of AC, is equal to twice the rectangle BC.BD; but twice the rectangle BC. BA is to twice the rectangle BC.BD; [that is, to the difference between the sum of the squares of AB, BC, and the square of AC]; as AB to BD; that is, by Prop. 1. as radius to the sine of BAD, (which is the complement of the angle ABC,) that is, as radius to the cosine of ABC. COR. If I be assumed as radius, then cos ABC = AB'+BC-AC2 2AB.BC * PROP. VII. FIG. 11. In any plane triangle, the rectangle contained by half the perimeter, and its excess above the base, is to the rectangle contained by the excesses of half the perimeter above the other two adjacent sides, as the square of the radius is to the square of the tangent of half the contained angle. Let ABC be a plane triangle, of which the base is BC, and let P denote the perimeter of the triangle; then }Р(¿P—BC): (†P—ÂB). (¿P—AC) : : rad2 : tan2 §BAC. Bisect the angles ABC, BAC by the straight lines BG, AG, meeting each other in G; join GC; and having produced the sides AB, AC, bisect the exterior angles CBH, BCL, by the straight lines BK, CK, meeting AG in K ; and from the points G, K, draw GD, GE, GF; KH, KL, KM, perpendiculars to the sides AB, AC, BC. Then, because G is the centre of the circle inscribed in the triangle ABC, (4. 4.) GD, GE, GF are equal, and AD is equal to AE, and BD to BF; also, since the angles at E and Fare right angles, the square of GC is equal to the squares of GE and EC together, or to the squares of GF and FC together; but the square of GE is equal to the square of GF; therefore, the square of EC is equal to the square of FC; and EC is equal to FC: and (8. 1.) the angle ECG is equal to FCG. For a like reason, KH, KL, KM are equal, and BH is equal to BM, and AH to AL, because the angles HBM, HAL are bisected by the straight lines BK, KA; and because in the two triangles KCL, KCM, the sides LK, KM are equal, KC is common to both, and KLC, KMC are right angles, CL is equal to CM. Since, therefore, BM is equal to BH, and CM to CL, BC is equal to BH and CL together: to these equals add A B and AC together, then AB, AC and BC together are equal to AH and AL together, but AH is equal to AL; wherefore each of them is equal to half the perimeter of the triangle ABC. But since AD is equal to AE, and BD to BF, and also CE to CF; AB and FC together are equal to half the perimeter of the triangle, to which AH or AL was shown to be equal; from these equals take away AB, which is common, and the remainder FC is equal to the remainder BH. In the same manner it may be demonstrated, that BF is equal to CL; and since the points B, D, G, F are in a circle, the angle DGF is equal to the exterior and opposite angle FBH, (22. 3); wherefore their halves BGD, HBK are also equal to one another. The right angled triangles BGD, HBK are therefore equiangular, and DG: DB:: BH: HK; therefore the rectangle DG.HK is equal to DB.BH. BAC. But AH: HK:: rad: (tan HAK=) tan. And AD: DG:: rad: (tan DAG=) tan. therefore, AH.AD: HK.DG :: rad2 : tan2 substitution, AH.AD: DB.BH:: rad: tan2 BAC; that is, P(P-BC): (P—AC). (¿P—AB): : rad: tan2 BAC. Q. E. D. * PROP. VIII. FIG. 11. In a plane triangle, the rectangle contained by any two adjacent sides, is to the rectangle contained by half the perimeter, and its excess above the base, as the square of the radius to the square of the cosine of half the contained angle. In the plane triangle ABC, if the perimeter be denoted by P, then will AB.AC:{P({P—BC): rad: cos BAC. For, the same construction being made as in the preceding proposition, in the right angled triangles AHK, ADG, AK: AH::rad: (cos HAK=) cos BAC, And AG: AD:: rad: (cos DAG=) cos BAC; therefore, AK.AG:AH.AD:: rad2: cos2 BAC. But since the straight lines BG, BK bisect the adjacent angles ABC, CBH, the angle GBK is a right angle, and for a like reason, GCK is a right angle; the points B, G, C, K are therefore in a circle, and the angles BKG, (or BKA,) and BCG in the same segment are equal; but BCG has been shown to be equal to ACG, and BAK is, by construction, equal to GAC; therefore the triangles ABK, AGC are equiangular; consequently AK: AB:: AC: AG, and the rectangle AK.AG is equal to AB.AC. Wherefore, AB.AC: AH.AD: : rad2: cos2 BAC; that is, AB.AC: P(P-BC): : rad2: cos2 BAC. Q. E. D. * PROP. IX. FIG. 11. In a plane triangle, the rectangle contained by any two adjacent sides is to the rectangle contained by the excesses of half the perimeter above those sides, as the square of the radius to the square of the sine of half the contained angle. In the plane triangle ABC, if the perimeter be denoted by |