Bisect AB in D, and draw BE at right angles to it, so that BE be equal to C; and having joined DE, from the centre D at the distance DE describe a circle meeting AB produced in G. And because AB is bisected in D, and produced to G, (6. 2.) AG.GB+DB2 und z DG2 DE2.

But (37. 1.) DE2=DB2+BE2, there

fore AG.GB + = DB + BE3, and

E

Exa

is equal to the square of C. by batsood ins (08) sertions 900 of has 997b seit eurood bis(3) Cub All e oa 4 or 80 and qu wough adr el PROP. XXX. PROB.ogong ons anil instan

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.

D

Upon AB describe (Prob. 18. 1.) the square BC, and produce CA to D, so that the rectangle CD.DA may be equal to the square CB (29. 6.). Take AE equal to AD, and complete the rectangle DF under DC and AE, or under DC and DA. Then, because the rectangle CD.DA is equal to the square CB, the rectangle DF is equal to CB. Take away the common part CE from each, and the remainder FB is equal to the remainder DE. But FB is the rectangle contained by FE and EB, that is, A by AB and BE; and DE is the square upon AE; therefore AE is a mean proportional between AB and BE (17. 6.), or AB is to AE as AE to EB. But AB is greater than AE; wherefore AE is greater than EB (14. 5.): Therefore the straight line AB is cut in extreme and mean ratio in E (def. 3. 6.).

Otherwise.

E

B

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 (9. 4.): Then because the rectangle AB.BC is equal to the square of AC, as BA to AC, so is AC to CB (17. 6.); Therefore AB is cut in extreme and mean ratio in C (def. 3. 6.).

A

C

B

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.

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. 20. 6.) Therefore, as CB to BD,

so is the figure upon CB to the similar and 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 together to BC, so are the figures upon BA and on AC, together, to the figure

B

D

upon BC (24. 5.); therefore the figures on BA, and on AC, are together equal to that on BC; and they are similar figures.

pa

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 rallel to one another; their remaining sides shall be in a straight line. Let ABC, DCE be two triangles which have 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, and AC to DE; BC and CE are in a straight line.

D

Because AB is parallel to DC, and the straight line AC meets them, the alternate angles BAC, ACD are equal (21. 1.); for the same reason, the angle CDE is equal to the angle ACD; wherefore also BAC is equal to CDE: And because the triangles ABC, 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

N

B

E

the angle DCE: And the angle BAC was proved to be equal to ACD: Therefore the whole angle AČE is equal to 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 (25. 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 (7. 1.) BC and CE are in a straight line.

PROP. XXXIII. THEOR.

T

In equal circles, angles, whether at the centres or circumferences, have the same ratio which the arcs, 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 BAC, EDF at their circumferences; as the arc BC to the arc 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 arcs CK, KL, each equal to BC, and any number whatever FM, MN each equal to EF; and join GK, GL, HM, ÍN. Because the arcs BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal (26.3.): Therefore, what multiple soever the arc BL is of the arc BC, the same multiple is the angle BGL of the angle BGC: For the same reason, whatever multiple the arc EN is of the arc EF the same multiple is the angle EHN of the angle EHF. But if the arc BL, be equal to the arc EN, the angle BGL is also equal (26. 3.) to the angle EHN; or if the arc BL be greater than EN, likewise the angle BGL is greater than EHN and if less, less: There being then four magnitudes, the two arcs, BC, EF, and the two angles BGC, EHF, and of the arc BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the arc BL, and the angle BGL; and of the arc EF, and of the angle EHF, any equimultiples whatever, viz. the arc EN, and the angle EHN: And it has been proved, that if the arc BL be greater than EN; the angle BGL is greater than EHN; and if equal, equal; and if less, less; As therefore, the arc BC to the arc EF, so (def. 5. 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 arc BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the arcs 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 also contain equal angles; the base BC is equal (1. 1.) to the base CK, and the triangle GBC to the triangle GCK: And because the arc BC is equal to the arc 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 (26.3.); and the segment BXC is therefore similar to the segment COK (def. 9. 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; and in the same manner, the sectors EHF, FHM, MHN, may be proved equal to one another: Therefore, what multiple soever the arc BL is of the arc BC, the same multiple is the sector BGL of the sector BGC. For the same reason, whatever multiple the arc EN is of EF, the same multiple is the sector EHN of the sector EHF; Now if the arc BL be equal to EN, the sector BGL is equal to the sector

EHN; and if the arc 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 arcs BC, EF, and the two sectors BGC, EHF, and of the arc BC, and sector BGC, the arc BL and the sector BGL are any equimultiples whatever; and of the arc EF, and sector EHF, the arc EN and sector EHN, are any equimultiples whatever; and it has been proved, that if the arc BL be greater than EN, the sector BGL is greater than the sector EHN; if equal, equal; and if less, less; therefore (def. 5. 5.) as the arc BC, is to the arc EF, so is the sector BGC to the sector EHF.

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 (Prob. 7. 3.) ACB about the triangle, and produce AD 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 BA: AD: : EA: (4. 6.) AC, and consequently, BA.AC= (16. 6.) AD.AE=ED.DA (3. 2.) +DA2. But ED. DA=BD.DC, therefore BA.AC = BD.DC +DA2.

B

E

ען

C

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.

Describe (Prob. 7. 3.) the circle ACB about the triangle, and draw its diameter AE, and join EC; Because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle AEC, in the same segment (21. 3.); the triangles ABD, AEC are equiangular Therefore, as (4. 6.) BA to AĎ, so is EA to AC: and consequently the rectangle BA.AC is equal (16. 6.) to the rectangle EA.AD.

B

E

A

D