PROP. XXX. PROB. To cut a given straight line in extreme and mean Let AB be the given straight line; it is required to cut it in extreme and mean proportion. Divide AB in the point C, so that the rectangle contained by AB, BC may be equal* #11. 2. #17. 6. to the square of AC; A с в AB BC is equal to the square of AC; BA: AC :: AC: CB; therefore AB is cut in extreme and #3 Def. 6. mean proportion in C.* Which was to be done. PROP. XXXI. THEOR. In right angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar and similarly described figures upon the sides containing the right angle. Let ABC be a right angled triangle, having the right angle BAC: the rectilineal figure described upon BC shall be equal to the similar and similarly described figures upon BA, AC. 112.1. #8. 6. #4. 6. Draw the perpendiculart AD: therefore, because in the right angled triangle ABC, AD is drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC are similar* to the whole triangle ABC, and to one another: and because the triangle ABC is similar to ABD, CB: BA :: BA: BD; and because these three straight lines are proportionals, as the first is t #2 Cor. 20.6. the third, so is the figure upon the first to the similar and similarly described figure* upon the second; therefore as CB: BD :: the figure upon CB to the similar and similarly described figure upon BA: and inversely, as DB to BC, so is the figure upon BA to that upon BC: for a similar reason, as DC to CB, so is the figure upon B CA to that upon CB: therefore as BD #8.5. *20.5. and DC together to BC,* so are the figures upon BA, AC to that upon BC: but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described figures on BA, AC. Wherefore, in right angled triangles, &c. #7.5. *20. 6. Cor. 1. +13. 5. *47. 1. Otherwise : Q. E. D. Fig. on AB fig. on AC :: square on AB: square on AC; therefore† fig. on AB + fig. on AC: fig. on AC square on BC: square on AC. But fig. on BC: fig. on AC :: square on BC: square on AC; hence fig. on BC: fig. on AC :: fig. on AB+fig. on AC: fig. on AC. Therefore the consequents being equal, the antecedents are equal;* Cor. 1. that is, the figure on BC is equal to the sum of the similar figures on AB and AC. #9. 5. PROP. XXXII. THEOR. If two triangles which have two sides of the one proportional to two sides of the other, be joined at one angle so as to have their homologous sides parallel to one another; the remaining sides shall be in a straight line. Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to +1 Ax. #6. 6. +2 Ax. D is equal to the angle ACD; wherefore also BAC is equal to CDE: and because the triangles ABC, DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular* to DCE: therefore the angle ABC is equal to the angle DCE: and the angle BAC was proved to be equal to ACD; therefore the whole angle ACE is equal to the two angles ABC, BAC: add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, #32. 1. BAC, ACB; but ABC, BAC, ACB are equal* to two right angles; therefore also the angles ACE, ACB are equal to two right angles: and since at the point C in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore* BC and CE are in a straight line. Where#14. 1. fore, if two triangles, &c. Q. E. D. PROP. XXXIII. THEOR. In equal circles, angles, whether at the centres or circumferences, are to each other as the circumferences on which they stand are to one another: so also are the sectors. Let ABC, DEF be equal circles; and at their centres the angles BGC, EHF, and the angles BAC, EDF, at their circumferences: as the arc BC to the arc EF, so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF. Take any number of arcs CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF: and draw GK, GL, HM, HN. Because the arcs BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal: therefore what multiple soever the arc #27.3. *27.3. BL is of the arc BC, the same multiple is the angle BGL of the angle BGC: for a similar reason, whatever multiple the arc EN is of the arc EF, the same multiple is the angle EHN of the angle EHF: and if the arc BL be equal to the arc EN, the angle BGL is also equal to the angle EHIN; and if the arc BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less: therefore since there are four magnitudes, the two arcs BC, EF and the two angles BGC, EHF, and that of the arc BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the arc BL, and the angle BGL; and of the arc EF, and of the angle EHF, any equimultiples whatever, viz. the arc EN, and the angle EHN; and since it has been proved, that if the arc BL be greater than EN, the angle BGL is greater than EHN, and if equal, equal; and if less, less: therefore, as the arc BC to the arc EF, so is the angle BGC to Def. 4.P. the angle EHF: but as the angle BGC is to the #6. 5 Cor. angle EHF, so is the angle BAC to the angle EDF: for each is double* of each; therefore, as 148. *20. 3. By the angle BGL" must be understood in this demonstration the sum of the angles BGC, CGK, &c. For GL might happen to be the continuation of BG, or it might be confounded with GB; and in neither case would these lines form any angle. Similar remarks apply to the angle EHN," which is a short expression for the sum of the angles EHF, FHM, &c. the arc BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Also, as the arc BC to EF, so shall the sector BGC be to the sector EHF. Join BC, CK, and in the arcs BC, CK take any points X, O, and join BX, XC, CO, OK: then, because in the triangles GBC, GCK, the two sides BG, GC are equal to the two CG, GK, each to #4. 1. +3 Ax. each, and that they contain equal angles: the base BC is equal to the base CK, and the triangle GBC to the triangle GCK: and because the arc BC is equal to the arc CK, the remaining part of the whole circumference of the circle ABC is equalt to the remaining part of the whole circumference of the same circle: therefore the angle BXC is equal to the angle COK; and the segment BXC is therefore similar to the segment *x. Def. 3. COK; and they are upon equal straight lines, *27.3. BC, CK: but similar segments of circles upon *24. 3. equal straight lines, are equal to one another; therefore the segment BXC is equal to the segment COK: and the triangle BGC was proved to be equal to the triangle CGK; therefore the whole, the sector BGC is equal to the whole, the sector CGK: for a like reason, the sector KGL is equal to each of the sectors BGC, CGK: in a similar manner the sectors EHF, FHM, MHN may be proved equal to one another: therefore, what multiple soever the arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC; and for a like reason, whatever multiple the arc EN is of EF, the same multiple is the sector EHN of the sector EHF: and if the arc BL be equal to EN, the sector BGL is equal to the sector EHN; and if the arc BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, less since then there are four magnitudes, the two arcs |