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

a, 29.

a 29.

b 26.

C 4.

a 34.

b Ax. I.

c Ax. 6.

d Ax. 2.

€ 34.

f 29. 4.

h Ax. 3.

T

PRO P. XXXIV. THE OR.

HE oppofite fides and oppofite angles of every parallelogram are equal; and the diameter divides it into two

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 2. For the fame reason 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 base BC common: Therefore the triangles are equal; and BC bifects the parallelogram. Wherefore, &c.

PRO P. XXXV. and XXXVII. THE OR.

Parallelograms and triangles, conflitute upon the fame bafe,

and between the fame parallels, are equal between themfelves, viz. parallelogram to parallelogram, and triangle to triangle.

Let ABCD, EBCF, be two parallelograms (Fig. 2.) conftitute upon the fame bafe BC, and between the fame parallels BC, AF; the parallelograms ABCD, EBCF, are equal.

For, because AD, EF, are each equal to BC, they are equal to one another. If the point E coincide with D. (Fig. 1.) each of the parallelograms are double the triangle DBC; therefore equal to one another. If AD is lefs than AE, add DE to both; then the whole AE is equal to DF 4, DC to AB, and the angle FDC to EAB: Therefore the triangles FDC, EAB, are equal. Take DGE from both; the trapeziums, ADGB, FEGC, are equal". Add the triangle GBC to both; then the whole parallelogram ABCD is equal to the parallelogram

EBCFd. 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 equal: 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 fame bafe, and between the fame parallels; and every triangle conftitute upon the fame bafe, and betwixt the fame parallels, is half the rectangle.

PROP. XXXVI. and XXXVIII. THEOR.

Parallelograms and triangles, conflitute upon equal bafes, and

between the fame parallels, are equal to one another, viz. parallelogram to parallelogram, and triangle to triangle.

Let the parallelograms ABCD, EFGH, be conftitute upon the equal bafes BC, FG, and between the fame parallels AH, BG; the parallelogram ABCD will be equal to EFGH,

Ax. 7.

For, join EB, CH, the parallelograms AC, EG *, are each equal to the parallelogram EC2; therefore equal to one ano- a 35. ther b. Join AC, FH; then the triangles ABC, HGF, are e- b Ax. 1. qual. Wherefore, &c.

PRO P. XXXIX. THEOR.

QUAL triangles, conftitute upon the fame bafe, on the fame
fide, are between the fame parallels.

Let the equal triangles ABC, DBC, be conftitute upon the fame bafe, BC, on the fame fide; the right line AD, that joins their vertex, will be parallel to BC.

c 34. and Ax. 7.

If not, draw AE, parallel to BC; join EC; then the triangles ABC, EBC, are equal; but DBC is equal to ABC; a 35. therefore DBC, EBC, are equal, a part to the whole; which b HYP is impoffible. Therefore no line but AD is parallel to BC. Wherefore, &c.

PROP.

* Parallelograms arc expreffed by the letters at the oppofite angles.

BOOK I.

a 36.

b hyp.

a 35.

b 34.

a 10.

b 23. C 31. d 36.

C 41.

f Ax. 6.

PRO P. XL. THE OR.

QUAL triangles, conftitute upon equal bases, on the fame
Jide, are between the fame parallels.

Let ABC, DGE, be equal triangles, conflitute upon the equal bafes 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 b; therefore DGE is equal to FGE, a part to the whole; which is impoffible: Therefore AD is parallel to BE. Wherefore,

&c.

I'

PRO P. XLI. THE OR.

F a parallelogram and triangle be conftitute upon the fame base, and between the fame 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.

PRO P. XLII. PR O B.

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

T%

ba

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.

Bilect 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 equal; and ABC is double AEC; but the parallelogram EG is double the triangle EAC: Therefore the parallelogram EG is equal to the triangle ABC, and the angle FEC equal to D. Wherefore, &c.

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.

BOOK I.

For, because DB is the diameter of the parallelogram ABCD, the triangles ADB, DBC, are equal. For the fame reafon, 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.

T

angle.

PRO P. XLIV. PRO B.

apply a parallelogram to a given right line equal to a given
triangle, having an angle in it equal to a given right lined

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.

ს

Make the parallelogram FGBE equal to the triangle C, having the angle EBG equal to D; 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 b 29. than two right angles; then FE, HB, being produced, will meet in fome point; which let be K; through which draw KL 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 d; 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.

T

PRO P. XLV. PRO B.

O make a parallelogram equal to a given right lined figure,
having an angle in it equal to a given right lined angle.

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.