be the homologous side to EG, and LM, to GF: and because EF is equal to C and KM together, EF is greater than KM; therefore the straight line EG is greater than KL, and GF than LM: make GX equal to LK, and GO equal to LM, and complete the parallelogram XGOP: therefore XO is equal and similar to KM; but KM is similar to EF; wherefore also XO is similar to EF, and therefore XO and EF are about the same diameter (26. 6.): let GPB be their diameter, and complete the scheme: then because EF is equal to C and KM together, and XO a part of the one is equal to KM a part of the other, the remainder, viz. the gnomon ERO, is equal to the remainder C: and because OR is equal (34. 1.) to XS, by adding SR to each, the whole OB is equal to the whole XB: but XB is equal (36.1.) to TE, because the base AE is equal to the base EB; wherefore also TE is equal to OB: add XS to each, then the whole TS is equal to the whole, viz. to the gnomon ERO: but it has been proved that the gnomon ERO is equal to C, and therefore also TS is equal to C. Wherefore the parallelogram TS, equal to the given rectilineal figure C, is applied to the given straight line AB deficient by the parallelogram SR, similar to the given one D, because SR is similar to EF (24. 6.). Which was to be done.

PROP. XXIX. PROB.

Tog a iven straight line to apply a parallelogram equal to a given rectilineal figure, exceeding by a parallelogram similar to another given.*

Let AB be the given straight line, and C the given rectilineal figure to which the parallelogram to be applied is required to be equal, and D the parallelogram to which the excess of the one to be applied above that upon the given line is required to be similar. It is required to apply a parallelogram to the given straight line AB, which shall be equal to the figure C, exceeding by a parallelogram similar to D.

Divide AB into two equal parts in the point E, and upon EB describe (18. 6.) the parallelogram EL similar and similarly si

* See Note.
A a

A

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 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 b the figure upon CA to that upon

CB. Wherefore as BD and DC

D

together to BC, so are the figures upon BA, AC to that upon BC (24. 5.): but BD and DC together are equal to BC. There fore 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.

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, DCF 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, and 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 same reason, the angle CDE is equal to the angle ACD; wherefore also BAC is equal to CDE; and because the triangles ABC,

* See Note,

B

D

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 triangles 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 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 that their centres 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, 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 BI, be equal to the circumfer

• See Note

ence 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 GCK: 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 B XC 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 circumference 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 BGL is equal to the sector EHN: and if the circumference BL be greater than EN, the sector BGL is greatter 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, the 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 def. 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.