Book I. but thefe are alternate angles: Therefore AC is parallel to BD, and likewife equal. Wherefore, &c.

PROP. XXXIV. THEOR.

Teran are equals and the diameter divides it into two

HE oppofite fides and oppofite angles of every parallelagram

equal parts.

Let ABCD be a parallelogram, the oppofite fides AB, CD; AC, BD, are equal; the angle CAB equal to BDC, and ACD to ABD; and the diameter BC bifects it.

For, because AB is parallel to CD, and BC falls upon them, the angle ABC is equal to BCD. For the fame reafon ACB is equal to CBD; therefore the two angles ABC, ACB, in the triangle ABC, are equal to the two angles CBD, BCD, in the triangle BCD; and the fide BC common to both: Therefore the two fides AC, AB, of the one triangle, are equal to the two fides BD, DC, of the other, each to each; and the angle BAC equal to BDC, and ACD to ABD. Again, because the two fides AC, AB, are equal to the two fides BD, DC, each to each, and the angle CAB equal to BDC, the bafe BC common: Therefore the triangles are equal; and BC bifects the parallelogram. Wherefore, &c.

P

PRO P. XXXV. and XXXVII. T H E O R.

Arallelograms and triangles, confiitute upon the fame bafe, and between the fame parallels, are equal between themjelves, viz. parallelogram to parallelogram, and triangle to triangle.

EBCF d. If AD is greater than AE, take DE from both; then Book I. the remainder AE will be equal to DF", the triangle AEB to DFC. Add EBCD to both, then the parallelograms ABCD, d Ax. 2. EBCF, are equald: So, likewife, if the diameters AC, BF, beh Ax. 3. drawn, then the triangle ABC will be equal to FBC. Where- i 34. and fore, &c.

COR. Hence every parallelogram is equal to a right angled parallelogram, conftitute upon the same bafe, and between the fame parallels; and every triangle conftitute upon the fame bafe, and betwixt the fame parallels, is half the rectangle.

P4

PRO P. XXXVI. and XXXVIII. THEOR.

Arallelograms and triangles, constitute upon equal bases, and between the fame parallels, are equal to one another, viz. parallelogram to parallelogram, and triangle to triangle.

Ax. 7.

PROP. XL. THEOR.

TQUAL triangles, conftitute upon equal bases, on

the fame

Let ABC, DGE, be equal triangles, constitute upon the equal bases BC, GE, on the fame fide; then AD is parallel to BE. If not, draw AF parallel to BE; join FE; then the triangle ABC is equal to FGE; but DGE is equal to ABC; therefore DGE is equal to FGE, a part to the whole; which is impoffible: Therefore AD is parallel to BE. Wherefore,

&c.

PROP. XLI. THEOR.

Fa parallelogram and triangle be conftitute upon the fame bafc, and between the same parallels, the parallelogram will be double the triangle.

Let the parallelogram be ABCD, and triangle EBC, having the fame bafe BC, and be between the fame parallels AE, BC, the parallelogram ABCD is double the triangle EBC; join AC.

Then the triangle ABC is equal to EBC; but the parallelogram ABCD is double the triangle ABC; and therefore double EBC. Wherefore, &c.

T

PROP. XLII. PROB.

O conftitute a parallelogram equal to a given triangle, having an angle in it equal to a given right lined angle.

Let the given triangle be ABC, and right lined angle D, it is required to conftitute a parallelogram equal to the given triangle ABC, having an angle in it equal to D.

Bifect BC in Ea; make an angle CEF equal to Db; through A draw AG parallel to CE; and through C draw CG parallel to EF; join AE; then the triangles ABE, AEC, are equald; and ABC is double AEC; but the parallelogram EG is double the triangle EAC: Therefore the parallelogram EG is equal to the triangle ABCf, and the angle FEC equal to D. Wherefore, &c.

Ν

PRO P. XLIII.

THEOR.

Book I.

IN every parallelogram, the complements that stand about the diameter are equal to one another.

Let ABCD be a parallelogram; BD its diameter; the parts of which BK, KD, the diameters of the parallelograms HKFD, EBGK; the remaining parallelograms AEKH, KGCF, its complements, are equal to one another.

For, because DB is the diameter of the parallelogram ABCD, the triangles ADB, DBC, are equal. For the fame reason, a 34. the triangle HKD is equal to DFK, and EBK to BKG; where

fore the triangles HKD and EKB are equal to DFK, BKG, b. b Ax. 2. Take HKD, ĚKB, from ADB, and DFK, BKG, from DBC, c Ax 3. there remains AEKH equal to KGCF. Wherefore, &c.

PRO P. XLIV. PR O B.

Tatthe range in it equal to a given right lined

O apply a parallelogram to a given right line equal to a given

angle.

It is required, upon the given right line AB, to make a parallelogram equal to a given triangle C, having an angle in it equal to a given angle D.

b 29.

Make the parallelogram FGBE equal to the triangle C, having the angle EBG equal to Da; put BE in a right line with a 42. AB; and produce FG to H; through A draw AH parallel to GB, or FE; join HB. Now, because the angles EFH, FHA, are equal to two right angles, the angles EFH, FHB, are lefs than two right angles; then FE, HB, being produced, will meet in fome point; which let be K; through which draw KLc 17, Cor. parallel to FH; and produce GB, HA, to M, L ; wherefore FHLK is a parallelogram, whofe diameter is HK ; and whofe complements FGBE, BALM, are equal 4; but FGBE, was d 43. made equal to C; and the angle EBG equal to D; therefore BALM is equal to C, and the angle ABM equal to D. e 15. Wherefore, &c.

PRO P. XLV. PRO B.

10 make a parallelogram equal to a given right lined figure,

to a

It

Book I. It is required to make a parallolegram equal to a given right lined figure ABCD, having an angle in it equal to the right lined angle E: Join DB, and make the parallelogram FH equal to the triangle ABD2; the angle FKH equal to E. Upon the right line GH make GM equal to DCB, and the angle GHM equal to Eb; then FM is the parallelogram equal to

a 42.

44.

C 29.

ABCD.

For, because FH is a parallelogram, the angles FKH, GHK, are equal to two right angles; but the angles GHM, FKH, are each equal to the angle E; therefore equal to one another. Add GHK to both, then the angles GHM, GHK, are equal to FKH, GHK, that is, equal to two right angles; therefore KHM is a right line. For the fame reafon FGL is a right lined; but FK, LM, are each parallel to GH; therefore pae 30. and rallel to one another. Wherefore FM is a parallelogram^equal to the right lined figure ABCD, and an angle FKM equal to E. Wherefore, &c.

d 14.

conftruct.

a II.

b 3.

C 35.

d 34.

COR. Hence a parallelogram may be made equal to a given right lined figure of any number of fides; for a parallelogram can be made equal to any triangle upon any given right line.

PRO P. XLVI. PRO B.

To deferibe a square upon a given right line.

It is required to defcribe a fquare upon the given right line AB.

From the point A, in the given right line AB, draw the perpendicular AC; cut off AD equal to AB b; through D draw DE parallel to AB, and BE parallel to AD; then ADEB is a parallelogram, the oppofite fides of which are equal; that is, e by conftr. DE equal to AB, and BE to AD; but AD is equal to AB; therefore the four fides are equal to one another. But the angles ADE, BAD, are equal to two right angles; and BAD is a right angle; therefore ADE is likewife a right angle; but the oppofite angles of every parallelogram are equal": Def. 30. Therefore ADEB is a fquares. Wherefore, &c.

£ 29.

PRO P. XLVII. THE OR.

N every right angled triangle the Square defcribed upon the fide fubtending the right angle is equal to the fquares of the fides containing the right angle.