Divide AB in the point C, fo that the rectangle contained Book VI. by AB, BC be equal to the square of AC 8.

Then, because the rectangle AB, BC is equal A C B

to the fquare of AC, as BA to AC, fo is AC

g II. 2.

to CB: Therefore AB is cut in extreme and mean ratio in Cf. h 17.6. Which was to be done.

IN

PROP. XXXI. THEOR.

N right angled triangles, the rectilineal figure defcribed upon the fide oppofite to the right angle, is equal to the fimilar, and fimilarly described figures upon the fides containing the right angle.

Let ABC be a right angled triangle, having the right angle BAC: The rectilineal figure described upon BC is equal to the fimilar and fimilarly defcribed figures upon BA, AC.

f3. Def. 6.

Draw the perpendicular AD; therefore, because in the a 12. 1. right angled triangle ABC, AD is drawn from the right angle

at A perpendicular to the base BC, the triangles ABD, ADC

D

C 4. 6.

d 2. Cor. 20.6.

are fimilar to the whole triangle ABC, and to one another b: b 8.6. and because the triangle ABC is fimilar to ADB, as CB to BA, fo is AB to BD; and because these three ftraight lines are proportionals, as the first to the third, fo is the figure upon the first to the fimilar and fimilarly described fi- B gure upon the second d; therefore, as CB to BD, fo is the figure upon CB to the fimilar and fimilarly defcribed figure upon BA: and inversely, as DB to BC, fo is the figure upon BA toe B. 5. that upon BC: For the fame reafon, as DC to CB, fo is the figure upon CA to that upon CB. Wherefore, as BD and DC together to BC, fo are the figures upon BA, AC to that upon BC : But BD and DC together are equal to f 24 5. BC; therefore the figure described on BC is equal g to the fimilar and fimilarly defcribed figures on BA, AC. Wherefore, &c. Q. E. D.

g A.

5.

BOOK VI.

a 31. I.

IF

PROP. XXXII. THEOR.

F two triangles which have two fides of the one proportional to two fides of the other, be joined at one angle, fo as to have their homologous fides parallel to one another; the remaining fides fhall be in a ftraight line.

Let ABC, DCE be two triangles which have the two fides BA, AC proportional to the two CD, DE, viz. BA to AC, as CD to DE; and let AB be parallel to DC, and AC to DE, BC and CE are in a straight line.

Draw EF parallel to AB or CD, and let it meet AC prob34. I. duced in F; therefore CDEF is a parallelogram, and CD is therefore equal to EF, and DE to

C 29. I.

b

N equal circles, angles, whether at the centres or circumferences, have the fame ratio which the arches on which they ftand have to one another: fo also have 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 arch BC to the arch EF, fo is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the fector EHF.

Take any number of arches CK, KL, each equal to BC, fo that BL be greater than EF; and join GK, GL: And because the arches BC, CK, KL are all equal, the angles BGC, CGK, a 27. 3. KGL are alfo all equal; therefore the arch BL is the same

multiple of the arch BC, that the angle BGL is of the angle BGC. Take, in like manner, the arch EM the leaft multiple

of

A

D

M

L

G

H

a 27. 3.

of the arch EF, that is greater than the arch BL, and join Book VI. HM; and it may be proved as before, that the angle EHM is the fame multiple of the angle EHF: confequently the angle FHM contains EHF the fame number of times that the arch FM contains EF: and if the arch BL be equal to FM, the angle BGL is alfo equal to FHM: If not, the arch BL is greater than FM, but lefs than EM the next greater multiple of EF; and the angle BGL is

a

B

K

C

E

F

greater than FHM, but lefs than EHM: therefore each of the arches BL, FM contains EF the fame number of times, and each of the angles BGL, FHM contains EHF the fame number of times: But the arch FM and the angle FHM contain EF and EHF equally; therefore the arch BL and angle BGL contain EF and EHF equally: and BL and BGL are any equimultiples of BC and BGC; as many times, therefore, as any multiple of BC contains EF, fo many times does the fame multiple of the angle BGC contain EHF; therefore, as the arch BC is to the arch EF, fo is the angle BGC to the angle EHF. But b5. Def. 5. as the angle BGC to the angle EHF, fo is the angle BAC to EDF, for each is double of each: Therefore, as the arch BC d 20. 3. to the arch EF, fo is the angle BGC to EHF, and the angle BAC to the angle EDF.

C

Alfo, as the arch BC to EF, fo is the fector BGC to the fector EHF.

Join BC, CK, and in the arches BC, CK take any points. X, O, and join BX, XC, CO, OK: Then, because in the triangles GBC, GCK

the two fides BG, GC are equal to the two CG, GK, and that they contain equal angles; the bafe BC is equal to CK, and the triangle BGC to CGK: and because

e

the arch BC is equal

to CK; if each of them be taken from the whole circumference,

the remainder BAKC is equal to the remainder CBAK; there

C 15. 5.

e 4. I.

fore the angle BXC is equal to the angle COK; and the feg- a 27. 3.

ment

BOOK VI. ment BXC is therefore fimilar to the segment COK: and the are upon equal bases BC, CK; therefore they are equal 8 to one f11.Def.3. another: and the triangle BGC is equal to the triangle CGK; 8 24. 3. therefore the whole fector BGC is equal to the whole sector CGK. For the fame reason, the fector KGL is equal to each of the sectors BGC, CGK: Therefore the sector BGL is the fame multiple of the fector BGC, that the arch BL is of the arch BC. In the fame manner, it may be proved, that the sector EHM is the fame multiple of the sector EHF, that the arch EM is of the arch EF; and that, if the arch BL be equal to the arch FM, the sector BGL is equal to the sector FHM; and if BL be not equal to FM, it is greater than it, but less than EM; therefore the sector BGL is greater than the sector FHM, but less than the sector EHM: Wherefore the arch BL and the sector BGL contain the arch EF and sector EHF the fame number of times that the arch FM and sector FHM contain them, that is, equally: as many times, therefore, as any multiple of the arch BC contains the arch EF, fo many times does the fame multiple of the fector BGC contain the sector b5.Def. 5. EHF; therefore, as the arch BC to the arch EF, fo is the fector BGC to the sector EHF. Wherefore, &c. Q. E. D.

a 5. 4.

PROP. B. THEOR.

F an angle of a triangle be bifected by a straight line, which alfo cuts the bafe; the rectangle contained by the fides of the triangle is equal to the rectangle contained by the fegments of the base, together with the fquare of the ftraight line bifecting the angle.

Let ABC be a triangle, and let the angle BAC be bifected by the ftraight line AD; the rectangle BA, AC is equal to the rectangle BD, DC, together with

the fquare of AD.

Defcribe the circle ACB about the triangle, and produce AD to E, and join EC: Then, because the angle BAD is equal to CAE, and b 21. 3. the angle ABD to the angle AEC ↳, for they are in the fame fegment ABEC; the triangles ABD, AEC are equiangular: therefore, as BA € 4. 6. to AD, fo is EA to AC; and confequently the rectangle BA, AC

is

d

is equal to the rectangle EA, AD, that is, to the rectangle Book VI. ED, DA, together with the fquare of AD: But the rectangle n ED, DA is equal to the rectangle BD, DC; therefore the d 16. 6. rectangle BA, AC is equal to the rectangle BD, DC, together with the fquare of AD. Wherefore, &c. Q. E. D.

PROP. C. THEOR.

[F from any angle of a triangle, a ftraight line be drawn perpendicular to the base; the rectangle contained by the fides of the triangle, is equal to the rectangle contained by the perpendicular and the diameter of the circle defcribed about the triangle.

Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC; the rectangle BA, AC is equal to the rectangle contained by AD and the diameter of the circle described

AEC are equiangular: Therefore,

E

e 3.2.

f

35.3.

A

as d BA to AD, fo is EA to AC; and confequently the rectangle d 2.6. BA, AC is equal to the rectangle EA, AD. If, therefore, e 16. 6. &c. Q. E. D.

THE

PROP. D. THEOR.

HE rectangle contained by the diagonals of a quadrilateral infcribed in a circle, is equal to both the rectangles contained by its oppofite fides.

Let ABCD be a quadrilateral inscribed in a circle, and draw the diagonals AC, BD; the rectangle AC, BD is equal to the two rectangles AB, CD and AD, BC.

If the angle ABD be equal to DBC, let AC, BD meet in E: But, if they be not equal, let ABD be the greater; and make a the angle ABE equal to the angle DBC; therefore, if the

angle

a 23. I.