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Book I. equal to the angle BCD; therefore the triangle ABC is equal

c to the triangle BCD, and the diameter BC divides the paralleloc 4. 1.

gram ACDB into two equal parts. Q. E. D.

PROP. XXXV. THEOR.

See N.

PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another.

See the Let the parallelograms ABCD, EBCF be upon the same base 2d and 3d BC, and between the same parallels AF, BC; the parallelogram figures. ABCD shall be equal to the parallelogram EBCF.

If the sides AD, DF of the parallelograms ABCD, DBCF opposite to A

D

F the base BC be terminated in the same

point D, it is plain that each of the a 34. 1. parallelograms is double a of the trian

gle BDC; and they are therefore equal
to one another.
But, if the sides AD, EF, opposite B

с
to the base BC of the parallelograms
ABCD, EBCF, be not terminated in the same point, then, be-

cause ABCD is a parallelogram, AD is equal a to BC; for the b 1. Ax. same reason, EF is equal to BC; wherefore AD is equal b to

EF, and DE is common; therefore the whole, or the remainc 2. or 3. der AE, is equal to the whole, or the remainder DF; AB also Ax. is equal to DC; and the two EA, AB are therefore equal to the

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two FD, DC, each to each; and the exterior angle FDC is d 19. 1. equal d to the interior EAB; therefore the base EB is equal to 4. 1. 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 theref3. Ax. fore are equalf, that is, the parallelogram ABCD is equal to the

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

Book I.

PROP. XXXVI. THEOR..

PARALLELOGRAMS upon equal bases, and between the same parallels, are equal to one another.

Let ABCD, EFGH be A

D E

H parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram ABCD is equal to EFGH. Join BE, CH; and be. B

С F

G cause BC is equal to FG, and FG to a EH, BC is equal to EH ; a 34 1. and they are parallels, and joined towards the same parts by the straight lines BE, CH: but straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel b; therefore EB, CH are both equal and pa-b 33. 1. rallel, and EBCH is a parallelogram; and it is equal to ABCD, c 35. 1. because it is upon the same base BC, and between the same 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.

PROP. XXXVII. THEOR.

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 same parallels

E
A D

F
AD, BC : the triangle ABC
is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B draw a BE parallel to CA;

a 31.1. and through C draw CF parallel to BD: therefore each

B

с of the figures EBCA, DBCF is a parallelogram; and EBCA is equal b to DBCF, because

b 35. 1 they are upon the same base BC, and between the same parallels BC, EF ; and the triangle ABC is the half of the parallelo

F

Book I. gram EBCA, because the diameter AB bisects e it; and the tri.

angle DBC is the half of the parallelogram DBCF, because the c 34. 1. diameter DC bisects it: but the halves of equal things are d 7. Ax. equal d; therefore the triangle ABC is equal to the triangle

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

PROP. XXXVIII. THEOR.

TRIANGLES 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 same parallels BF, AD : the triangle ABC is equal to the triangle DEF.

Produce AD both ways to the points G, H, and through B a 31. 1. draw BG parallel a to CA, and through F draw FH parallel to ED: then each of

G

A
D

H
the figures GBCA,
DEFH is a parallel-
ogram,

and they b 36. 1. are equal b to one

another, because they
are upon equal base's
BC, EF, and be.
tween the same pa-

B
C E

F c 34. 1. rallels BF, GH; and the triangle ABC is the half c of the pa.

rallelogram GBCĄ, because the diameter AB bişects it; and the triangle DEF is the half of the parallelogram DEFH, because

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

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

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PROP. XXXIX. THEOR.

EQUAL triangles upon the same base, and upon the same side of it, are between the same parallels.

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

Join AD; AD is parallel to BC; for, if it is not, through the 31. 1. point A draw : AE parallel to BC, and join EC: the triangle

ABC is equal b to the triangle EBC, because it is upon the same Book 1. base BC, and between the same parallels BC, AE: but the triangle ABC is A

D b 37. 1. equal to the triangle BDC ; therefore also the triangle BDC is equal to the

E tộiangle EBC, the greater to the less, which is impossible: therefore AE is not parallel to BC. In the same manner, it can be demonstrated that no other line but AD is parallel to BC; AD is B

с therefore parallel to it. Wherefore, equal triangles, &c. Q. E. D.

PROP. XL. THEOR.

EQUAL triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels.

Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight

A

D line BF, and towards the same parts; they are between the same parallels.

Join AD; AD is parallel to BC: for, if it is not,

a 31. L through A draw a AG parallel to BF, and join GF:

B

C E the triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and between the same 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 impossible : 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.

Fb 38. 1.

b

PROP. XLI. THEOR.

IF a parallelogram and triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle.

Book I. Let the parallelogram ABCD and the triangle EBC be upon

the same base BC, and between the same parallels BC, AE ; the
parallelogram ABCD is double of the
triangle EBC.

A

D E Join AC; then the triangle ABC a 37. 1. is equal a to the triangle EBC, because

they are upon the same base BC, and

between the same parallels BC, AE. b 34. 1. But the parallelogram ABCD is double b

of the triangle ABC, because the diame-
ter AC divides it into two equal parts;
wherefore ABCD is also double of the

B

с triangle EBC. Therefore, if a parallelogram, &c. Q. E. D.

PROP. XLII. PROB.

TO describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles

equal to D. a 10.1. Bisect a BC in E, join AE, and at the point E in the straight b 23. 1. line EC make b the angle CEF equal to D; and through A draw c 31. 1. CAG parallel to EC, and through

A F G
C draw CG c parallel to EF:
therefore FECG is a parallelo-
gram: and because BE is equal

to EC, the triangle ABE is liked 38. 1. wise equal d to the triangle AEC,

D
since they are upon equal bases
BE, EC, and between the same
parallels BC, AG: therefore the
triangle ABC is double of the

в ЕС triangle AEC: and the parale 41. 1. lelogram FECG is likewise double e of the triangle AEC, be

cause it is upon the same base, and between the same parallels : therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D. Wherefore there has been described a parallelograr

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