« PreviousContinue »
because the parallelogram GH is greater than
* 26.6. their diameter FX,
LM and complete the
E scheme. Therefore,
BIO since GH is equal to EL and C together,
N P X and that GH is equal to MN; MN is equal 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* to the parallelo- *36. 1. gram NB, that
to BM:* add NO to each ; *43. 1. 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 figure C, exceeding by the parallelogram PO, which is similar to D, because PO is similar* to EL. Which was to be done. *24.6.
PROP. XXX. PROB.
Upon AB describe* the square BC, and to 46. 1. AC * apply the parallelogram CD, equal to BC, * 29. 6. exceeding by the figure AD similar to BC:
then, since BC is a square,
therefore also AD is a square : and because BC is equal to CD, by taking the common part CE from each, the remainder BF is equal to the remainder AD: and these figures are equiangular, therefore their sides about the
D * 14.6 equal angles are reciprocally* pro
portional: therefore, as FE to
ED, so AE to EB: but FE is A * 31. 1. equal * to AC, that is, tot AB; +30 Def. and ED is equal to AE; therefore as BA to AE, so is AE to EB:
С F but AB is greater than AE ; *14.5. wherefore AE is greater than EB : therefore
the straight line AB is cut in extreme and mean **Def.6. ratio in E.* Which was to be done.
Otherwise, Let AB be the given straight line; it is required to cut it extreme and mean ratio.
Divide AB in the point C, so that the rectangle contained by AB, BC, may be equal
of AC: A CB then, because the rectangle AB, BC is equal to the square of AC; as BĂ to AC, so is AC to
CB : * therefore AB is cut in extreme and mean #3Def.6. ratio in C.* Which was to be done.
PROP. XXXI. THEOR.
described upon the side opposite to the right
Let-ABC be a right angled triangle, having the right angle BAC: the rectilineal figure described upon BC shall be 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 *8.6 another: 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 is to the third, so is the figure upon the first to the similar and similarly described figure * upon the second: therefore as *2 Cor. CB to BD, so is the figure upon CB to the simi- 20.6. lar and similarly described figure upon BẢ: and inversely, * as DB to BC, so is the figure upon BA to that upon
BC: for the same Bf reason, as DC to CB, so is the figure upon CA to that upon CB : therefore as BD and DC together to
so are the figures upon BA, AČ to that *24. 5. upon BC: but BD and DC together are equal to BC; therefore the figure described on BC is equal to the similar and similarly described * A.5. figures on BA, AC. Wherefore, in right angled triangles, &c. Q.E.D.
* B. 5.
PROP. XXXII. THEOR.
proportional to two sides of the other, be joined
Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC, as
A CD to DE; and let AB be parallel to DC, and AC to DE: BC and CE shall be in a straight line.
Because AD is parallel to DC, and the straight line
* 29. 1. AC meets them, the alternate angles * BAC,
ACD are equal; for the same reason, the angle
CDE is equal to the angle ACD; where+1 Ax. fore also BAC is equal t 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 DĚ, the triangle ABC is equiangular * 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 +2 Ax. is equal t to the two angles ABC, BAC: add the common
angle ACB, then the angles ACE, ACB are equal to the angles ABC, BĂC, ACB : but * 32. 1. ABC, BAC, ACB are equal * to two right angles;
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 * BC and CE are in a straight line. Wherefore, if two triangles, &c Q. E. Ď.
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 : 80 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 circumference BC to the circumference EF, so shall the angle BGC be 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
ference BL be greater than EN, the angle BGL is greater than EHN; and if equal, equal; and if less, less : therefore as the circumference BC to the circumference EF, so * is the angle BGC #5Def.5. to the angle EHF: but as the angle BGC is to the angle EHF, so is * the angle BAC to the *15.5. angle EDF: for each is double * of each ; there- * 20.3. fore, 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 shall