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fimilar and alike fituate to D, and equal to the excess by which Book VI. EF exceeds Cb; then EF is equal to C, and KLMN together; therefore KLMN is lefs than EF; and, because they are fimilar, the a 18. fide GF is greater than LM, and GE than LK; make GO equal 6 25. to LM, and GX to LK; and compleat the parallelogram GP, which will be fimilar to, and about the fame diameter with EF; let this diameter GB be drawn, and produce XP to R, and OP to S; then TS will be equal to C, and wanting in figure by SR, which is fimilar to Dd: Which was to be done.

C 26.

d 27.

PRO P. XXIX. PROB.

To apply a parallelogram upon a given right line, equal to a given right lined figure, exceeding by a parallelogram fimilar to another given parallelogram.

Upon the given right line AB, it is required to apply a parallelogram equal to the given right lined figure C, exceeding by a parallelogram fimilar to the given parallelogram D.

Bifect AB in E; upon EB defcribe the parallelogram EL, fimilar and alike fituate to Da; and the parallelogram GH equal a 18. to C and EL together ", and fimilar and alike fituate to Da; b 251 let KH be a fide homologous to FL, and KG to FE; then, because the parallelogram GH is greater than the parallelogram EL, the right line KH is greater than FL, and KG than FE: Produce FL and FE to M and N, fo that FM be equal to KH, and FN to KG; compleat the parallelogram NM; then MN is equal and fimilar to GH; but GH is fimilar to EL; therefore MN is fimilar to EL ; therefore EL is about the fame diameter c 21. with MN d; let FX, their diameter, be drawn, and defcribe the d 26. figure: Then, fince GH is equal to EL and C together, as also to MN; therefore MN is equal to EL and C together. Take KL, which is common, from both; then the gnomon MPE is equal to C; and, because the parallelograms AN, NB, are equal, AN is equal to LO f; and, if BX be added, AX is equal to the gnomon MPE; therefore AX is equal to C. Where- Ax. 1. J. fore, &c.

e 36. I.

f 43. and

PRO P. XXX. PR O B.

O cut a given right line into extreme and mean ratio,
N

T.

It

Book VI.

a 46. I.

b 29.

C 14.

d 11.2.

It is required to cut the given right line AB into extreme and

mean ratio.

Upon AB defcribe the fquare BCa, and to AC apply the parallelogram CD, equal to the fquare BC, exceeding by the fi gure AD, fimilar to BC; but BC is a fquare; therefore AD is alfo a fquare. From the equal parallelograms BC, CD, take a way the common parallelogram CE; then the remainder BF will be equal to AD; but BF is equal to AD; therefore FE is to ED as AE is to EB; that is, AB is to AE as AE is to EB: Or, let AB be cut in E, fo that the rectangle under AB, BE, be equal to the fquare of AE d. Wherefore, &c.

PRO P. XXXI. THE OR.

IA N every right angled triangle, any figure described upon the fide fubtending the right angle, is equal to the two fimilar figures defcribed upon the fides containing the right angle.

Let ABC be the right angled triangle, the figure described on BC, fubtending the right angle, is equal to the two fimilar figures deferibed on BA, AC; for, from the point A, let fall the perpendicular AD; then the triangle ABC is divided into the two fimilar triangles ADB, ADC; then, because the triangle ABC is fimilar to the triangle ABD, CB is to BA as BA is to BD, and CB is to BD as the figure defcribed on CB is to b Cor. 20. the fimilar figure described on BA o. For the fame reason, as

a

BC is to CD, fo is the figure described on BC to the fimilar one defcribed on AC: Wherefore, as BC is to BD, and DC together, fo is the figure defcribed on BC to the two fimilar figures 24. 5. defcribed on BA, AC, together; but BC is equal to BD, and DC together; therefore the figure defcribed on BC is equal to the two fimilar figures defcribed on BA, AC. Wherefore, &c.

PRO P. XXXII. T HE OR.

IF two triangles having two fides proportional to two fides, be fo compounded or fet together at one angle, that their homologous fides be parallel; then the other fides of these triangles will be in one right line.

If the triangles ABC, DCE be fo placed at the point C, that the fide DE be parallel to AC, and DC to AB; then BÇÈ will be a right line.

For, becaufe the homologous fides AB, DC, are parallel, and Book VI. AC falls upon them, the alternate angles BAC, ACD, are e-" qual; for the fame reafon, CDE is equal to ACD; then, a 29. 1. fince the two triangles BAC, CDE, have the angles at A and D equal, and the fides about them proportional, viz. BA to AC as CD to DE, the triangles are equiangular, viz. theb 6. angle ABC equal to DCE, and ACB to DEC; but the angle ACD is proved equal to BAC; therefore, the whole angle ACE is equal to the two angles ABC, BAC. Add the common angle ACB to both, then the two angles ACE, ACB, are equal to the three angles ABC, ACB, BAC; that is, equal to two right angles; therefore BCE is one right line, Wherefore,c 32. 1. &c.

d 14. I.

PRO P. XXXIII. THE OR.

IN equal circles, the angles are in the fame proportion to one another as the circumferences on which they fand, whether the angles be at the centres or the circumference: fo likewife are fectors, as being at the centres.

Let ABC, DEF, be equal circles, and the angles BGC, EHF, at the centres G, H; and BAC, EDF, angles at their circumferences; then the angle BGC will be to the angle EHF as the circumference BC is to the circumference EF; and likewife the angle BAC to the angle EDF, and the sector BGC to the fector EHF, as the circumference BC to EF. For, take any number of circumferences, as CK, KL, each equal to BC; and any number of circumferences, as FM, MN, each equal to EF; join GK, GL, HM, HN; then, because the circumferences BC, CK, KL, are equal, the angles BGC, CGK, KGL, are likewife equal; therefore, BL is the fame multiple a 27. 3. of BC, that the angle BGL is of the angle BGC; for the fame reafon, EN is the fame multiple of EF, that EHN is of EHF; therefore, if the circumference BL be equal to the circumference EN, the angle BGL is equal to the angle EHN, if greater greater, and if lefs lefs; therefore, as BC is to EF, fo is BGC to EHF; and fo is BAC to EDF. Again, as the b def. 5. 5. circumference BC is to EF, fo is the fector BGC to the fector 15. 5.and EHF; for, join BC, CK, EF, FM, and affume the points X, O, in the circumference BC, CK, and join BX, XC, CO, OK, then, becaufe BG, GC, are equal to CG, GK, and contain equal angles, the bafe BC is equal to the bafe CK d ̧d 4. I. and the triangles equal; and, because the right line BC is e

qual

C

20.3.

e 28. 3

Book VI.qual to the right line CK, the circumference BXC is equal to the circumference COK: therefore the angle BXC is equal to the angle COK: Therefore, the fegments BXC, COK, are equal f 24. 3. and and fimilar f; but the triangles BGC, CGK, are equal; theredef. II. 3. fore the whole fector BGCX is equal to the whole fector CGKO; in the fame manner the fectors EHF, FHM, are proved equal; therefore BK is the fame multiple of BC, that BGK is of BGCX; and EM the fame multiple of EF that the sector EHMF is of the sector EHF; therefore, if the circumference BK is equal to the circumference EM, the fector BGK is equal to the fector EHM, if greater, greater, and if lefs, lefs; thereb def. 5. 5. fore, as BC is to EF, fo is the fector BGCX to the fector EHF. Wherefore, &c.

COR. I. An angle at the centre of a circle is to four right. angles, as the arch on which it ftands is to the whole circumference; for, as the angle BAC is to a right angle, fo is the arch BC to a quadrant, the confequents quadrupled; then BAC is to four right angles as BC is to the whole circumference.

COR. II. The arches IL, BC, of unequal circles, which fubtend equal angles, whether at the centres or circumferences, are fimilar: For IL is to the whole circumference ILE as the angle IAL, or BAC, is to four right angles; and fo is the arch BC to the whole circumference BCF; therefore the arches IL, BC, are fimilar.

COR. III. Two femidiameters AB, AC, cut off fimilar arches IL, BC, from concentric circumferences.

THE

f

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