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

a 46. I.

b 29.

C 14.

d 11.2.

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 away the common parallelogram ĈE; 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 AEd. Wherefore, &c.

PROP. XXXI. T HEOR.

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

Let ABC be the right angled triangle, the figure defcribed on BC, fubtending the right angle, is equal to the two fimilar figures defcribed 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 defcribed on BA b. For the fame reason, as

a

S.

BC is to CD, fo is the figure defcribed on BC to the similar one defcribed on AC: Wherefore, as BC is to BD, and DC together, fo is the figure described 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.

I

PROP. XXXII. THEOR.

F 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 BCE will be a right line.

For;

For, becaufe the homologous fides AB, DC, are parallel, and Book VI. AC falls upon them, the alternate angles BAC, ACD, are equal; for the fame reason, 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. theь 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 lined. Wherefore,c 32. I. &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 stand, whether the angles be at the centres or the circumference; fo likewife are fcators, 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 fector 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. 3of 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 4, 4. 1. and the triangles equal; and, because the right line BC is e

qual

C

20.3.

€ 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; theredet. 11. 3. fore the whole fector BGCX is equal to the whole sector 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 fector EHF; therefore, if the circumference BK is equal to the circumference EM, the fector BGK is equal to the sector EHM, if greater, greater, and if lefs, lefs; thereb def. 5. 5. fore, as BC is to EF, fo is the sector BGCX to the sector 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 so 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.