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AD: And these figures are equian

Otherwise, gular; therefore their sides about the Let AB be the given straight line ; equal angles are reciprocally propor- it is required to cut it in extreme and tional: (11. 6.) Wherefore, as FÈ to

mean ratio. ED, so AE to EB : But FE is equal Divide AB in the point C, so that to AC, (34. 1.) that is to AB; and the rectangle contained by AB, BC ED is equal to AE:

be equal to the square of AC: (11. 2.) Therefore as BA to AE,

Then, because the rect. so is AE to AB: But

angle AB, BC is equal

А C B AB is greater than AE;

to the square of AC, as wherefore AE is greater

BA to AC so is AC to CB: (17. 6.) than EB:(14. 5.) There

Therefore AB is cut in the extreme fore the straight line AB

and mean ratio in C. (3. def. 6.) is cut in extreme and

Which was to be done. mean ratio in E. (3. def. 6.) Which was to be done.

PROP. XXXI. THEOREM.

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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, CB to BD, so is having the right angle BAC: The the figure upon rectilineal figure described upon BC CB to the similar is equal to the similar, and similarly and similarly dedescribed figures upon BA, AC. scribed figure

Draw the perpendicular AD; there- upon BA: And, fore, because in the right angled tri- inversely, (B. 5.) B angle ABC, AD is drawn from the as DB to BC, so right angle at A perpendicular to the is the figure upon BA to that upon base BC, the triangles ABD, ADC BC: For the same reason, as DC to are similar to the whole triangle ABC, CB, so is the figure upon ĆA to that and to one another, (8. 6.) and be upon CB. Wherefore as BD and cause the triangle ABC is similar to DC together to BC, so are the figures ABD, as CB to BA, so is BA to BD; upon BA, AC to that upon BC: (t. 6.) and because these three (24. 5.) But BD and DC together are straight lines are proportionals, as equal to BC. Therefore the figure the first to the third, so is the figure described on BC is equal (A. 5.) to upon the first to the similar, and the similar and similarly described similarly described figure upon the figures on BA, AC. Wherefore, in second: (2 Cor. 20. 6.) Therefore as right angled triangles, &c. Q. E. D.

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PROP. XXXII. THEOREM.

If two triangles which have two sides of the one proportional to iwo

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 proportional to the two CD, DE, viz. which have the two sides BA, AC BA to AC, as CD to DE; ard let AB

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be parallel to DC, and AC to DE, the angle BAC was proved to be equal BC and CE are in a straight line. to ACD: Therefore the whole angle

Because AB is parallel to DC, and ACE is equal to the two angles ABC, the straight line ACA

BAC; add the common angle ACB, meets them, the al

then the angles ACE, ACB are equal ternate angles BAC

to the angles ABC, BAC, ACB: But ACD are equal; (29

ABC, BAC, ACB are equal to two 1.) for the same rea.

right angles ; (32. 1.) therefore also son, the angle CDEB

the angles ACE, ACB are equal to is equal to the angle ACD; where- two right angles : And since at the fore also BAC is equal to CDE: And point C, in the straight line AC, the because the triangles ABC, DCE have two straight lines BC, CE, which are one angle at A equal to one at D, on the opposite sides of it, make the and the sides about these angles pro- adjacent angles ACE, ACB equal to portionals, viz. BA to AC, as CD to two right angles; therefore (14. 1.) DE, the triangle ABC is equiangular BC and CE are in a straight line. (6. 6.) to DCE: Therefore the angle Wherefore, if two triangles, &c. Q. ABC is equal to the angle DCE: And E. D.

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PROP. XXXIII. THEOREM.

In equal circles, angles, whether at the centres or circumferences, have

the same ratio which the circumferences on which they stand have to one another: So also have the sectors.

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Let ABC, DEF be equal circles; EHN: and if the circumference BL and at their centres the angles BGC, be greater than EN, likewise the EHF, and the angles BAC, EDF at angle BGL is greater than EHN; their circumferences ; as their circum- and if less, less: there being then ference BC to the circumference EF, four magnitudes, the two circumferso is the angle BGC to the angle EHF, ence BC, EF, and the two angles and the angle BAC to the angle EDF; BGC, EHF; of the circumference BC, and also the sector BGC to the sector and of the angle BGC, have been EHF.

taken any equimultiples whatever, Take any number of circumferences viz. the circumference BL, and the CK, KL, each equal to BC, and any angle BGL; and of the circumfernumber whatever FM, MN, each ence EF, and of the angle EHF, any equal to EF: And join GK, GL, HM, HN. Because the circumferences BC, CK, KL, are all equal, the angles BGC, CGK, KGL are also all equal : (27. 3.) Therefore what multiple so

N ever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC:

E F For the same reason, whatever mula tiple the circumference EN is of the equimultiples whatever, viz. the circircumference EF, the same multiple cumference EN, and the angle EHN: is the angle EHN of the angle EHF: And it has been proved, that, if the And if the circumference BL be equal circumference BL begreater than EN, to the circumference EN, the angle the angle BGL is greater than EHN; BGL is also equal (27. 3.) to the angle and if equal, equal; and if less, lens:

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As therefore the circumference BC to manner, the sectors EHF, FHM, the circumference EF, so (5. def. 5.) MHN may be proved equal to one is the angle BGC to the angle EHF: another: Therefore, what multiple But as the angle BGC is to the angle soever the circumference BL is of the EHF, so is (15. 5.) the angle BAC to circumference BC, the same multiple the angle EDF; for each is double of is the sector BGL of the sector BGC: each (20. 3.): Therefore, as the cir- for the same reason, whatever mulcumference BC is to EF, so is the tiple the circumference EN is of EF, angle BGC to the angle EHF, and the the same multiple is the sector EHN angle BAC to the angle EDF. of the sector EHF: And if the cir.

Also, as the circumference BC to cumference BL be equal to EN, the EF, so is the sector BGC to the sector EHF. Join BC, CX, and in the circumferences BC, CK take any points X, 0, and join BX, XC, CO,

N OK: Then, because in the triangles GBC, GCK the two sides BG, GC

M are equal to the two CG, GK, and that they contain equal angles; the base BC is equal (4. 1.) to the base sector BGL is equal to the sector CK, and the triangle GBC to the tri- EHN; and if the circumference BL angle GCK: And because the circum. be greater than EN, the sector BGL ference BC is equal to the circumfere is greater than the sector EHN; and ence CK, the remaining part of the if less, less: Since then, there are whole circumference of the circle four magnitudes, the two circumferABC, is equal to the remaining part ences BC, EF, and the two sectors of the whole circumference of the BGC, EHF, and of the circumference same circle: Wherefore the angle BC, and sector BGC, the circumferBXC is equal to the angle COK (27. ence BL and sector BGL are any 3.); and the segment BXC is there- equimultiples whatever; and of the fore similar to the segment COK (11. circumference EF, and sector EHF, def. 3.); and they are upon equal the circumference EN, and sector straight lines BC, CK: But similar EHN, are any equimultiples whatsegments of circles upon equal straight ever; and that it has been proved, if lines, are equal (2+. 3.) to one ano- the circumference BL be greater than ther: Therefore the segment BXC is EN, the sector BGL is greater than equal to the segment COK ; And the the sector EIN; and if equal, equal; triangle BGC is equal to the triangle and if less, less. Therefore, (5. def. 5.) CGK; therefore the whole, the sec- as the circumference BC is to the cirtor BCG, is equal to the whole, the cumference EF, so is the sector BGC sector CGK: For the same reason, to the sector EHF. Wherefore, in the sector KGL is equal to each of equal circles, &c. Q. E. D. the sectors BGC, CGK: In the same

PROP. B. THEOREM.

If an angle of a triangle be bisected by a straight line which likewise

cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.

Let ABC be a triangle, and let straight line AD; the rectangle BA, the angle BAC be bisected by the AC is equal to the rectangle BD,

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DC, together with the square of AD. another: Therefore as BA to AD, 90 Describe the cir.

is (4. 6.) EA to AC, and consequently cle (5.4.)ACB about

the rectangle BA, AC is equal (16. 6.) the triangle, and

to the rectangle EA, AD, that is, (3. produce AD to the

2.) to the rectangle ED, DA, together circumference in E,

with the square of AD: But the rectand join EC: Then

angle ED, DA is equal to the rect. because the angle

angle (35.3.) BD, DC. Therefore the BAD is equal to

rectangle BA, AC is equal to the rectthe angle CAE, and the angle ABD angle BD, DC, together with the to the angle (21. 3.) AEC, for they square of AD. Wherefore, if an are in the saine segment, the triangles angle, &c. Q. E. D ABD, AEC are equiangular to one

PROP C. THEOREM.

If from an angle of a triangle a straight line be drawn perpendicular to

the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a triangle, and AD the and the angle ABD perpendicular from the angle A to the to the angle AEC in base BC, the rectangle BA, AC is the same segment; equal to the rectangle contained by (31. S.) the triangles AD, and the diameter of the circle ABD, AEC are equidescribed about the triangle. angular: Therefore

Describe (5. 4.) the circle ACB a- as (4.6.) BA to AD, bout the triangle, and draw its dia- so is EA to AC; and meter AE, and join EC: Because the consequently the rectangle BA, AC is right angle BDA is equal (31. 3.) equal (16. 6.) to the rectangle EA, AD. to the angle ECA in a semicircle, If therefore from an angle, &c. Q.E.D.

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PROB. D. THEOREM.

The rectangle contained by the diagonals of a quadrilateral inscribed in

a circle, is equal to both the rectangles contained by its opposite sides.

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Let ABCD be any quadrilateral in- the angle BCE, scribed in a circle, and join AC, BD; because they are the rectangle contained by AC, BD is in the same segequal to the two rectangles contained ment ; therefore by AB, CD, and by AD, BC • : the triangle ABD

Make the angle ABE eyual to the is eqniangular to angle DBC , add to each of these the the triangle BCE: common angle EBD, then the angle Wherefore, (4. 6.) ABD is equal to the angle EBC; and as BC is to CE so is BD to DA; and the angle BDA is equal (21. 3.) to consequently the rectangle BC, AD is

A

This is a Lemma of A. Ptolomæus, in page 9 of his joyahn aurratu.

equal(16.6.) to the rectangle BD,CE: AE: But the rectangle BC, AD has Again, because the angle ABE is equal been shewn equal to the rectangle BD, to the angle DBC, and the angle (21. CE; therefore the whole rectangle AC, 3.) BAE to the angle BDC, the triangle BD (4. 6.) is equal to the rectangle ARE is equiangular to the triangle AB, DC, together with the rectangle BCD: As therefore BA to AE, so is AD, BC. Therefore the rectangle, BD to DC; wherefore the rectangle &c. Q. E. D. BA, DC is equal to the rectangle BD,

BOOK XI.

DEFINITIONS.

I.

at right angles to it, one upon one A solid is that which hath length, plane, and the other upon the other breadth, and thickness.

plane. II.

VII. That which bounds a solid is a su- Two planes are said to have the same, perficies.

er a like inclination to one another, III.

which two other planes have, when A straight line is perpendicular, or at the said angles of inclination are

right angles to a plane, when it equal to one another. makes right angles with every

VIII. straight line meeting it in that Parallel planes are such which do not plane.

meet one another though produced. IV.

IX. A plane is perpendicular to a plane, A solid angle is that which is made by

when the straight lines drawn in the meeting of more than two plane one of the planes perpendicularly angles, which are not in the same to the common section of the two plane, in one point. planes, are perpendicular to the

X. Other plane.

· The tenth definition is omitted for

reasons given in the notes.' The inclination of a straight line to a

XI. plane is the acute angle contained Similar solid figures are such as have by that straight line, and another all their solid angles equal each to drawn from the point in which the each, and which are contained by first line meets the plane, to the the same number of similar planes. point in which a perpendicular to

XII. the plane, drawn from any point of A pyramid is a solid figure contained the first line above the plane, meets by planes that are constituted bethe same plane.

twixt one plane and one point above VI.

it in which they meet. The inclination of a plane to a plane а

XIII. is the acute angle contained by two A prism is a solid figure contained by straight lines drawn from any the plane figures, of which two that are same point of their common section opposite are equal, similiar, and par

V.

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