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Upon AB describe the square BC, and to AC apply the paral- Book VI. lelogram CD equal to BC exceeding by the figure AD similar to BC b. but BC is a square, therefore also
a. 45 1. AD is a square. and because BC is equal
b. 29.6. to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD. and these figures are A equiangular, therefore their sides about the equal angles are reciprocally proportionale.
C. 14. 6. wherefore as FE to ED, so AE to EB. but FE is equal to AC4, that is to AB; and
d. 34. I. ED is equal to AE. therefore as BA to C
F AE, fo is AE to EB. but AB is greater than AE; wherefore AE is greater than EBo, therefore the straight e. 14. S. line AB is cut in extreme and mean ratio in Ef. Which was to be f. 3.17f. 6. done.
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%. then because the rectangle AB, BC is equal to the CB fquare of AC, as BA to AC, fo is AC to CBh.
h. 17.6. therefore AB is cut in extreme and mean ratio in Cf. Which was to be done.
PROP. XXXI. THEOR, IN right angled triangles the rectilineal figure describeci See N.
upon the side opposite to the right angle, is equal to the similar, and similarly described figures upon the tides containing the right angle.
Let ABC be a right angled triangle, having the right angle B4C. the rectilineal figure described upon BC is equal to the fimilar 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*. and because the triangle a. 8.6,
Book VI. ABC is fimilar to ABD, as CB to BA, fo is B A to BD b. ant le
ncause these three straight lines are proportionals, as the first to the b. 4. 6. third, fo is the figure upon the first to the similar, and fimi.dly dc. C. 2. Cor. fcribed figure upon the secondo.
therefore as CB to BD, so is
figure upon BA. and, inverse.
с 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 t. 24. 5. BC, so are the figures upon BA, AC to that upon BC°. but BD
and DC together are equal to BC. Therefore the figure described f. A. s. on BC is equal to the similar and similarly described figures on BA,
AC. Wherefore in right angled triangles, &c. Q. E. D.
PROP. XXXII. THEOR.
portional to two sides of the other, be joined at one angle so as to have their homologous fides parallel to one another; the remaining fides shall be in a straight line.
Let ABC, DCE be two triangles which have the two sides BA,
Because AB is parallel to
D a. 29.1. cqual o ; for the same reason
the angle CDE is equal to
C E and because the triangles ABC, DCE have one angle at A equal to one at D, and the fides about these angles proportionals, viz. B A to AC, as CD to DE,
the triangle ABC is equiangular b to DCE. therefore the angle Book VI. ABC is equal to the angle DCE. and the angle BAC was proved m to be equal to ACD, therefore the whole angle ACE is equal to b. 6.6. 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"; therefore also the C. 37. 5. angles ACE, ACB are equal to two right angles. and since at the point C in the Itraight 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 a BC and CE are in a straight d. 14. 1. line. Wherefore if two triangles, &c. Q. E. D.
N equal circles, angles whether at the centers or cir. See N.
cumferences 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 centers the angles BGC, EHF, and the angles BAC, 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, A
E F 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", there- 2. 27. 3. fore 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 the circumference EN is of
Book VI. the circumference EF, the same multiple is the angle EHN of the mangle EHF. and if the circumference BL be equal to the circumfea. 27. 3. rence EN, the angle BGL is also equal a to the angle EHN; and if
the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if lefs, 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,
C 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, cqual; and if lefs, lefs. as therefore the circumference BC b. s. Dcf. s. to the circumference EF, fo b is the angle BGC to the angle EHF. C. 15. 5. but as the angle DGC is to the angle EHF, so is the angle BAC to d. 20. 3. the angle EDF, for each is double of eachd. therefore as the cir
cumference 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, fo 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 e. 4. 1. CG, GK, and that they contain equal angles; the base BC is equal
to the base CK, and the triangle GBC to the triangle GCK. and because the circumference BC is equal to the circumference CK, the rčinaining 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* ; and f. 11. Dcf.3. the fegment BXC is therefore similar to the segment COKf; and
they are upon equal straight lines BC, CK, but similar segments of g. 24. 3. circles upon equal straight lines, are equal é to one another. there
fore the segment BXC is equal to the segment COK. and the tri- Book VI.
M B XC E F rence 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 fector BGL are any equimultiples whatever; and of the circumference EF and 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 EŇ, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore b as the circumference BC is to the circumference EF, so is b. s. Def.s. the sector BGC to the sector EHF. Wherefore in equal circles, &c. Q. E. D.