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to EL and C: take away the common part EL: then the remainder, viz. the gnomon NOL, is equal to C. And because AE is equal to EB, the parallelogram AN is equal (36. 1.) to the parallelogram NB, that is, to BM (43. 1.). Add NO to each; therefore the whole, viz. the parallelogram AX is equal to the gnomon NOL. But the gnomon NOL is equal to C; therefore also AX is equal to C. Wherefore to the straight line AB there is applied the parallelogram AX equal to the given rectilineal C, 3. by the parallelogram PO, which is similar to D, because PO is similar to EL (24.6). Which was to be done.

PROP. XXX. PROB.

To cut a given straight line in extreme and mean ratio.

Let AB be the given straight line; it is required to cut it in extreme and mean ratio.

Upon AB describe (46. 1.) the square BC, and to AC apply the parallelogram CD equal to BC, exceeding by the figure AD similar to BC (29. 6.): but BC is a square, therefore |D also AD is a square; and because BC is equal to CD, by taking the common part CE from R. each, the remainder BF is equal to the remain- A der AD: and these figures are equiangular, therefore their sides about the equal angles are reciprocally proportional (14.6); wherefore as FE to ED, so AE to EB; but FE is equal to AC (34.1.), that is, to AB; and ED is equal to AE: therefore as BA to AE, so is AE to EB; but AB is greater than AE; where- C F fore AE is greater than EB (14. 5): therefore the straight line AB is cut in extreme and mean ratio in E (3. def. 6). Which was to be done.

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Otherwise.

Let AB be the given straight line; it is required to cut it in extreme and mean ratio.

Divide AB in the point C, so that the rectangle contained by AB, BC be equal to the square of AC (11. 2.), Then, -—— because the rectangle AB, BC is equal to the square A C B of AC, as BA to AC, so is AC to CB (17. 6.): therefore AB is cut in extreme and mean ratio in C (3. def. 6.). 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.*

* See Note.

Let ABC be a right angled triangle, having the right angle BAC; the rectilineal figure described upon BC is equal to the similar and similarly described figures upon BA, AC.

Draw the perpendicular 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 (8.6), and because the triangle ABC is similar to ADB, as CB to BA, so is BA to BD (4.6.); and because these three straight lines are proportionals, as the first to the third, so is the figure upon the first to the similar and similarly described figure upon the second (2 Cor.): therefore as CB to BD, so is the figure upon CD to the similar and A similarly described figure upon BA: and, inversely (B. 5), as DB to BC, so is the figure upon BA to that upon BC; for the same reason, as DC to CB, so is the figure upon CA to that upon CB. Wherefore as BD and DC B - C together to BC, so are the figures D upon BA, AC to that upon BC (24. 5): but BD and DC together are equal to BC. Therefore the figure described on BC is equal (A. 5.) to the similar and similarly described figures on BA, AC. Wherefore, in right angled triangles, &c. Q. E. D.

PROP. XXXII. THEOR. v

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 AC, as CD to DE; and let AB be parallel to DC, as AC to DE. BC and CE are in a straight line.

Because AB is parallel to DC, A and the straight line AC meets them, the alternate angles BAC,

ACD are equal (29. 1.); for the D same reason, the angle CDE is w N equal to the angle ACD; where- \ ~ fore also BAC is equal to CDE: and because the triangles ABC, B C E

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 (6. 6.) 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

* See Notes.

the two angles ABC, BAC; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB: but ABC, BAC, ACB are equal to two right angles (32. 1.); 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 (14. 1.) BC and CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D.

PROP. XXXIII. THEOR.

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.”

Let ABC, DEF be equal circles: and at their centres the angles BGC, EHF, and the angles BAD, EDF at their circumferences: as the circumference BC to the circumference EF, so is the angle BGC 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 circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each 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 soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC : for the same reason, whatever multiple of the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF: and if the circumference BL be equal to the circumference EN, the angle BGL is also equal (27. 3.) to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less: there being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF: of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BL, and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples whatever, viz. the circumference EN, and the angle EHN: and it has been proved that if the circumference BL be greater than EN, the angle BGL is greater than EHN: and if equal, equal; and if less, less: as therefore the circumference BC to the circumference EF, so (5. def. 5.) is the angle BGC to the angle EHF: but as the angle BGC is to the angle EHF, so is (15. 5.) the angle BAC to the angle EDF, for each is double of each (20. 3.): therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Also, as the circumference BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the circumferences 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, and that they contain equal angles; the base BC is equal (4.1.) to the base CK, and the triangle GBC to the triangle GUK : and because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle: wherefore the angle BXC is equal to the angle COK (27. 3.), and the segment BXC is therefore similar to the segment COK (11. def. 3.); and they are upon equal straight lines BC, CK: but similar segments of circles upon equal straight lines, are equal (24, 3) to one another: therefore the segment BXC is equal to the segment COK: and the triangle BGC is equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK: for the same reason, the sector KGL is equal to each of the sectors BGC, CGK: in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another: therefore, what multiple soever the circumference BL is of the circumference BC, the same multiple is the sector BGL of the sector BGC : for the same reason, whatever multiple the circumfeernce EN is of EF, the same multiple is the sector EHN of the sector EHF : and if the circumference BL be equal to EN, the sector BGE,

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is equal to the sector EHN: and if the circumference 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 circumferences BC, EF, and the two sectors BGC, EHF, and of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equal multiples whatever: and of the circumference EF, and

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sector EHF, the circumference EN and sector EHN are any equimultiples whatever; and that it has been proved, if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore (5. des. 5.), as the circumference BC is to the circumference EF, so is the sector BGC to the sector EHF. Wherefore, in equal circles, &c. Q. E. D.

PROP. B. THEOR.

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 the angle BAC be bisected by the straight line AD; the rectangle BA, AC is equal to the rectangle BD, DC together with the square of AD. Describe the circle (5. 4.) ACB about the triangle, and produce AD & A

to the circumference in E, and join EC: then because the angle BAD is equal to the angle CAE, and the angle ABD to the angle (21. 3.) AEC, for they are in the same segment: the triangles ABD, AEC, are equiangular to one another: therefore as BA to AD, so is (4.6.) EA to AC, and consequently the rectangle BA, AC is equal (16. 6.) to the rectangle EA, AD, that is (3. 2.), to the rectangle ED, DA, together with the square of AD: but the rectangle ED, DA, is equal to the rectangle (35. 3.) BD, DC. Therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, &c. Q. E. D.

PROP. C. THEOR.

IF from any 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 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 about the triangle.

* See Note. 20.

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