About B, as a centre, with BC, BA for distances, let two circles CD, EA be described, meeting BA, BC, in D, E ; since CAB is a right angle, BC being radius, AC is the sine of the angle ABC, by def. 4. and BA being radius, AC is the tangent, and BC the secant, of the angle ABC, by def. 6. 7. Cor. 1. Of the hypothenuse, a side, and an angle of a right-angled triangle, any two being given, the third is also given. Cor. 2. Of the two sides and an angle of a right-angled triangle, any two being given, the third is also given. PROPOSITION II. The sides of a plane triangle are to one another, as the sines of the angles opposite to them. In right-angled triangles, this Proposition is manifest from Prop. 1. for if the hypothenuse be made radius, the sides are the sincs of the angles opposite to them, and the radius is the • Cor. Def. sine of a right angle * which is opposite to the hypothenuse. 4. of this. In any oblique-angled triangle ABC, any two sides AB, AC will be to one another as the sines of the angles ACB, ABC, which are opposite to them. From C, B, draw CE, BD, perpendicular с Fig. 6. Fig.7. B E Cor. Hence of two sides, and two angles opposite to them, in a plane triangle, any three being given, the fourth is also given. D с D PROPOSITION III. In a plane triangle, the sum of any two sides is to their dif ference, as the tangent of half the sum of the angle at the base, to the tangent of half their difference. 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 A 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* is equal to the sum of 32. 1. 1 BAG the angles at the base, and the angle EFB at the circumference is equal to the half of E EAB at the centre *; therefore EFB is half * 20. 3. the sum of the angles at the base; but the Fig. 8. angle ACB * is equal to the angles CAD * 32. 1. 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 *, FB being radius, BE, BG are the tangents of the an- * 1. of this. gles 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 *, EC is to CF, as EB to BG ; that is, the sum of the 2. 6. 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. PROPOSITION IV. 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 perpendicnlar to BA; make AE, AF, each equal to AB, and AD to AC; join BE, BF, BD, and from D draw DG perpendicular And because BA is at right an Fig.9. F gles to EF, and EA, AB, AF are equal, cach of the angles EBA, ABF is half a right angle, and the whole EBF is a right angle; also * EB is • 4. 1. A The figures in the margin refer to the Elements of Euclid, unless otherwise expressed, upon BF. 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 parallel to GD, BG is to GF, as ED is to DF; that is, since ED is the sum of * 3. of this. the sides BA, AC, and FD their difference *, 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. PROPOSITION V. In any triangle, twice the rectangle contained by any two sides, is to the difference of the sum of the squares of these tro 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 ABC contained by any two sides BA, BC, is to the difference of the sum of the squares of BA, BC, and the square of the base AC, as Fig.10. B the radius to the cosine of B Fig 11. the angle ABC. From A, draw AD perpen dicular upon the opposite side • 12. & 13. BC, then * the difference of the sum of the squares of AB, BC, 2. and the square of the base AC, is equal to twice the rectangle CBD; but twice the rectangle CBA is to twice the rectangle CBD, that is to the difference of the sum of the squares of * 1. 6. AB, BC, and the square of AC * as AB to BD; that is, as * 1. of this. radius to the sine of BAD, which is the complement of the angle ABC; that is, as radius to the cosine of ABC. PROPOSITION VI. BC, the rectangle contained by half the perimeter, and the ceeds the other two sides AB, AC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. E D • 4. 4. G FIM H Let the angles BAC, ABC be bisected by the straight lines AG, BG; and producing the side AB, let the exterior angle CBH be bisected by the straight line BK, meeting AG in K; and from the points G, K, let there be drawn perpendicular upon the sides, the straight lines GD, GE, GF, KH, KL, KM. Since therefore G is the centre of the circle inscribed in the triangle ABC; B GD, GF, GE will be equal, and AD will be equal to AE, BD to BF, and CE to CF. In like manner KH, KL, KM will be equal, and BH will be equal to BM, and AH to AL, be- Fig.12. K cause the angles HBM, HAL are bisected by the straight lines BK, KA: and because in the triangles KCL, KCM, the sides LK, KM are equal, Kc is common, and KLC, KMC are right angles, CL will be equal to CM: since therefore BM is equal to BH, and CM to CL; BC will be equal to BH and CL together; and, adding AB and AO together, AB, AC, and BC will together be equal to AH and AL together : but AH, AL are equal : wherefore each of them is equal to half the perimeter of the triangle ABC: but since AD, AE are equal, and BD, BF, and also CE, CF; AB, together with FC, will be equal to half the perimeter of the triangle to which AH or AL was shewn to be equal; taking away therefore the common AB, the remainder FC will be equal to the remainder BH: in the same manner it is demonstrated, that BF is equal to CL: and since the points B, D, G, F are in a circle, the angle DGF will be equal to the exterior and opposite angle FBH *; wherefore their halves BGD, * 22. 3. HBK will be equal to one another : the right-angled triangles BGD, HBK will therefore be equiangular, and GD will be to BD, as BH to HK: and the rectangle contained by GD, HK, will be equal to the rectangle DBH or BFC: but since AH is to HK, as AD to DG, the rectangle HAD * will be to the rect- 22.6. angle contained by HK, DG, or the rectangle BFC, (as the square of AD is to the square of DG, that is,) as the square of the radius is to the square of the tangent of the angle DAG, FF that is, the half of BAC: but he is half the perimeter of the triangle ABC, and AD is the excess of the same above HD, that is, above the base BC; but BF or CL is the excess of HA or AL above the side AC; and FC, or HB, is the excess of the same HA above the side AB; therefore the rectangle contained by half the perimeter, and the excess of the same above the base, viz. the rectangle HAD, is to the rectangle contained by the straight lines by which the half of the perimeter exceeds the other two sides, that is, the rectangle BFC, as the square of the radius is to the square of the tangent of half the angle BAC opposite to the base. Q. E. D. PROPOSITION VII. In a plane triangle, the base is to the sum of the sides as the difference of the sides is to the sum or difference of the segments of the base made by the perpendicular upon it from the vertex, according as the square of the greater side is greater or less than the sum of the squares of the lesser side and the base. Tig 14. A G Let ABC be a plane triangle ; if from A the vertex be of AC the greater C G BD E E About A, as a centre, with AC the greater side for a distance, let a circle be described meeting AB produced in E, F, and CB in G: it is manifest, that FB is the sum, and BE the difference of the sides ; and since AD is perpendicular to GC; GD, CD will be equal; consequently GB will be equal to the sum or difference of the segments CD, BD, according as the perpendicular AD meets the base produced, or the base; that is (by Conv. 12. 13. 2.), according as the square of AC is greater or less than the sum of of AB, BC: but * the rectangle CBG is equal to the rectangle EBF; that is * BC is to BF, as BE is to BG; that is, the base is to the sum of the sides, as the difference of the squares . 35. 3. • 16. 6. |