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gles o, each to each, to which the equal sides are opposite. there- Book I. fore the angle ACB is equal to the angle CBD. and because the n straight line BC meets the two straight lines AC, BD and makes b. 4. I. the alternate angles ACB, CBD equal to one another, AC is parallel - to BD. and it was shewn to be equal to it. therefore c. 29. 1. straight lines, &c. Q. E. D.

PRO P. XXXIV. THEO R.

a. 29. I.

THE
HE opposite sides and angles of parallelograms are

equal to one another, and the diameter bisects them, that is, divides them into two equal parts.

N. B. A Parallelogram is a four sided figure of which the opposite sides are parallel. and the diameter is the straight line joining two of its opposite angles.

Let ABCD be a parallelogram, of which BC is a diameter. the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to

А

B one another, and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. wherefore the two triangles ABC, CBD have two angles ABC, BCA

С

D in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the other 5, viz. the side AB to the side CD, and AC to BD, and b. 26. 3. the angle BAC equal to the angle BDC. and because the angle ABC is equal to the angle BCD, and the angle CED to the angle ACB; the whole angle ABD is equal to the whole angle ACD. and the angle BAC has been shewn to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another. also, their diameter bifects them. for, AB being equal to CD, and BC common; the two AB, BC are equal to the Iwo DC, CB, sach to each ; and the angle ABC is equal to the

Book I. angle BCD; therefore the triangle ABC is equal to the triangle WBCD, and the diameter BC divides the parallelogram ACDB into

two equal parts. Q. E. D.

c. 4. 1.

See N.

See the

a. 34. 1.

PRO P. XXXV. THE O R. PARALLELOGRAMS upon the same base and between

the fame parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the same base ad and 3d

BC and between the fame parallels AF, BC. the parallelogiam Figures.

ABCD shall be equal to the parallelogram EBCF.

If the fides AD, DF of the parallelograms ABCD, DBCF opposite to the base BC, be terminated in the fame point D; it is plain that each of the parallelograms is double * of the A D F triangle BDC; and they are therefore equal to one another.

Butif the sides AD, EF opposite to the base BC of the parallelograms ABCD, EBCF be not terminated in the fame B point; then because ABCD is a parallelogram, AD is equal to

BC; for the same reason, EF is equal to BC; wherefore AD is b. 1. ds. equal b to EF ; and DE is common; therefore the whole, or the e... or 3. remainder, AE is equal to the whole, or the remainder DF;

AB alfo is equal to DC ; and the two EA, AB are therefore equal

A D E F A E DF

Ax

d. 29. 1.

B С

B C to the two FD, DC, each to each ; and the exterior angle FDC is equal to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal e to the triangle FDC. take the triangle FDC from the trapezium ABCF, and from the

same trapezium take the triangle EAB; the remainders therefore are f. 3. Ax equalf, that is, the parallelogram ABCD is equal to the parallelogram

EBCF. therefore parallelograms upon the same base, &c. Q. E.D.

C. 4. 1.

Book 1.

PRO P. XXXVI. THEOR.

PARALLELOGRAMS upon equal bases and between

the fame parallels, are equal to one another.

G 4.3

Let ABCD, EFGH be parallelograms upon equal bafcs BC, FG, and between the same parallels AH, BG; the paral- A D E H lelogram ABCD is equal to EFGH.

Join BE, CH; and becaufe BC is equal to FG, and FG to · EH, BC is

a. 34. di equal to EH ; and

they B C F are parallels, and joined towards the fame parts by the straight lines BE, CH. but straight lines which join equal and parallel straight lines towards the fame parts, are themselves equal and parallel b; therefore EB, CH are both equal and parallel, and b. 33. 5. EBCH is a parallelogram; and it is equal o to ABCD, because it C. 35.8. is upon the fame base BC, and between the fame parallels BC, AD. for the like reason the parallelogram EFGH is equal to the same EBCH. therefore also the parallelogram ABCD is equal to EFGH. Wherefore parallelograms, &c. Q. E. D.

PRO P. XXXVII. THEO R. TRIANGLES upon the same base, and between the

same parallels, are equal to one another. Let the triangles ABC, DBC be upon the same base BC and between the fame parallels E AD F AD, BC. the triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and thro' B draw · BE parallel to CA;

By :3. i. and thro' C draw CF parallel to BD. therefore each of the

B C figures EBCA, DBCF is a parallelogram ; and EBCA is equal o tob. 35. 3. DBCF, because they are upon the fame base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the pas

C

Book 1. rallelogram EBCA, because the diameter AB bisects it; and the

triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it. but the halves of equal things are equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

C. 34. I.

Ax.

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PRO P. XXXVIII. THE O R.

TRI

RIANGLES upon equal bases, and between the

same parallels, are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the fame parallels BF, AD. the triangle ABC is equal to the triangle DEF.

Produce AD both ways to the points G, H, and thro' B draw 2. 31. I. BG parallel • to CA, and thro' F draw FH parallel to ED. then

each of the figures
GBCA, DEFH is a
G А.

H parallelogram ; and b. 36. I.

they are equal b to
one another, because
they are upon equal
bases BC, EF and be-
tween the fame paral-

B CE F
lels DF, GI; and the triangle ABC is the half of the paral-
lelogram GBCA, because the diameter AB bisects it; and the
triangle DEF is the half c of the parallelogram DEFH, because

the diameter DF bisects it. but the halves of equal things are d. 7. Ax. equal ; therefore the triangle ABC is equal to the triangle

DEF. Wherefore triangles, &c. Q. E. D.

C. 34. 1.

PRO P. XXXIX. THEOR.

EQUAL triangles upon the same base, and

upon the

fame side of it, are between the fame parallels.

Let the equal triangles ABC, DBC be upon the same base BC, and upon the same side of it; they are between the fame parallels.

Join AD; AD is parallel to BC; for if it is not, thro' the point A draw. AE. parallel to BC, and join EC. the triangle ABC is

2. 31. I.

equal o to the triangle EBC, because it is upon the same base BC, Book I. and between the same parallels BC, AE.

А

D but the triangle ABC is equal to the tri

b. 37. 1, angie BDC; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impoflible. therefore AE is not parallel to BC. in the same manner it can be demonstrated B that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles upon, &c. Q. E. D.

PROP. XL.

THEOR.

EQUAL triangles upon equal bases, and towards the

fame parts, are between the fame parallels.

Let the equal triangles ABC, DEF be upon equal bases BC, EF, and towards the fame parts;

A D they are between the fame paralléls.

Join AD; AD is parallel to BC. for if it is not, thro'

a. 31.1 A draw · AG parallel to BF, and join GF. the triangle

b. 38. 1 ABC is equait to the triangleB CE F GEF, because they are upon equal bases BC, EF, and betweer the fame parallels BF, AG. but the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impoflible. therefore AG is not parallel to BF. and in the same manner it can be demonstrated that there is no other parallel to it but AD, AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D.

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IF
F a parallelogram and triangle be upon the same

base, and between the fame parallels ; the parallelogram shall be double of the triangle.

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