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Book I. but these are alternate angles : Therefore AC is parallel to

BD“, and likewise equal. Wherefore, &c.

a, 29.

PRO P. XXXIV. THEO R.

HE opposite sides and opposite angles of every parallelo

gran are equal; and the dian:eter divides it into two equal parts.

TI

2 29

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

For, because AB is parallel to CD, and BC falls upon them, the angle ABC is equal to BCDa. For the same reason ACB is equal to CBD; therefore the two angles ABC, ACB, in the triangle ABC, are equal to the two angles CBD, BCR, in the triangle BCD; and the side BC common to both: Therefore the two fides AC, AB, of the one triangle, are equal to the two sides BD, DC, of the other, each to cach; and the angle BAC equal to BDC, and ACD to ABDb. Again, because the two sides AC, AB, are equal to the two sides BD, DC, each to each, and the angle CAB equal to BDC, the base BC

Therefore the triangles arc equal; and BC bisects the parallelogram. Wherefore, &c.

b 26.

C 4.

common :

PRO P. XXXV. and XXXVII. THEO R.

Arallelograms and triangles, constitute upon the same base,

and bei-ween the fame parallels, are equal between themjelves, viz. parallelogram to parallelogram, and triangle to triangle.

a 34.

b Ax. I.

CAX. 6.

Let ABCD, EBCF, be two parallelograms (Fig. 2.) constitute upon the same base BC, and between the same parallels BC, AF; the parallelograms ABCD, EBCF, are equal.

For, because AD, EF, are cach equal to BC, they are equal to one another b. If the point E coincide with D. (Fig. 1.) each of the parallelograms are double the triangle DBC; therefore e

qual to one another. If AD is less than AE, add DE to d Ax, 2.

both; then the whole AE is equal to DF, DC to AB', and the angle FDC to LABf: Therefore the triangles FDC, EAB,

are equal 6. Take DGE from both; the trapeziums, ADGB, i Ax. 3. FEGC, are equal h. Add the triangle GBC to both; then thé whole parallelogram ABCD is equal to the parallelogram

C 34. f 29.

& 4.

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 m DFC. Add EBCD to both, then the parallelograms ABCD, Ax. 2. EBCF, are equald: So, likewise, if the diameters AC, BF, bé h Ax. 3. drawn, then the triangle ABC will be equal to FBCi, Where. i 34. and fore, &c.

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

AX. 7.

PRO P. XXXVI. and XXXVIII. THEOR.

P4

Arallelograms and triangles, constitute upon equal bases, and

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

Let the parallelograms ABCD, EFGH, be constitute upon the equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD will be equal to EFGH.

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

C 34. and

Ax. 7

PRO P. XXXIX. THE O R.

fame base, on the same

fide, are between the fame parallels.

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

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 impossible. Therefore no line but AD is parallel to BC. Wherefore, &c.

PROP

Parallelograms arc expressed by the letters at the opposite angles.

Book I.

PRO P. XL. THE O R.

QUAL triangles, constitute upon equal bases, on the same fide, are between the same parallels.

a 36,
b hyp:

Let ABC, DGE, be equal triangles, constitute upon the e. qual bases BC, GE, on the same lide; 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 6; therefore DGE is equal to FGE, a part to the whole; which is impoflible : Therefore AD is parallel to BE. Wherefore, &c.

PRO P. XLI. THEO R.

Fa parallelogram and triangle be constitute upon the same base,

and between the same parallels, the parallelogram will be double the triangle.

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Let the parallelogram be ABCD, and triangle EBC, having the same base BC, and be between the same 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 6; and therefore double EBC. Wherefore, &c.

à 35. b 340

P R O P. XLII. PRO B.

O constitute a parallelogram equal to a given triangle, ha

ving an angle in it equal to a given right lined angle.

,

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

Bilect BC in Eo; make an angle CET equal to Db; through A draw AG parallel to CE"; and through C draw CG parallel to EFC; join AL; then the triangles ABE, AEC, are equal d; and ABC is double ALC; but the parallelogram EG is doublé the triangle EAC : Therefore the parallelogram EG is equal to the triangle ABCf, and the angle FEC equal to D. Wherefore, &c.

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Book I.

PRO P. XLIII.

T H E O R.

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N every parallelogram, the complements that stand about the

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 same reason, a 34. the triangle HKD is equal to DFK, and EBK to BKG; wherefore the triangles HKD and EKB are equal to DFK, BKG, 5. b Ax. 2. Take HKD, EKB, from ADB, and DFK, BKG, from DBC, C Ax 3. there remains AEKH equal to KGCF. Wherefore, &c.

PRO P. XLIV. PRO B.

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

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

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

Make the parallelogram FGBE equal to the triangle C, having the angle EBG equal to D a; 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 less than two right angles; then FE, HB, being produced, will meet in some point; which let be K; through which draw KL parallel to FH; and produce GB, HA, to M, L; wherefore FHLK is a parallelogram, whose diameter is HK, and whose complements FGBE, BALM, are equal d; but TGBE, 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.

b 29.

C17:

Core

PRO P. XLV. PRO B.

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

having an angle in it equal to a given right lined angle.

It

a 42

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C 29

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 e. qual to the triangle ABD"; 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 ABCD.

For, because FH is a parallelogram, the angles FKH, GHK, are equal to two right angleso; 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 d. For the same reason 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 parallelograme.

qual to the right lined figure ABCD, and an angle FKM equal to E. Wherefore, &c.

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.

d 14

contruct.

P R O P. XLVI. PRO B.

Todefer

O describe a square upon a given right line.

b 3. с 31. d 36.

It is required to describe a square upon the given right line AB.

From the point A, in the given right line AB, draw the perpendicular Aca; cut off AD equal to AB b; through D draw DE parallel to ABC, and BE parallel tó AD, then ÅDEB is a

parallelogram, the opposite sides of which are equal d; that is, e by confr. DE equal to AB, and BE to AD; but AD is equal to AB €;

therefore the four sides are equal to one another. But the f 29.

angles DE, BAD, are equal to two right angles; and BAD is a right angle; therefore ADE is likewise a right angle ;

but the opposite angles of every parallelogram are equal 4: & Def. 30. Therefore ADEB is a squares. Wherefore, &c.

PRO P. XLVII. THE O R.

IN
Nevery right angled triangle the Square described upon the side

subtending the right angle is equal to the squares of the sides
taining the right angle.

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