PA

PROP. XXXVI. THEOR.

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

Let ABCD, EFGH be parallelograms upon equal bases BC, FG,

and between the same pa

rallels AH, BG; the paral- A

lelogram ABCD is equal
to EFGH.

Join BE, CH; and be

cause BC is equal to FG,
and FG to EH, BC is
B
equal to EH; and they

DE

CF

H

G

Book I.

a. 34. 1

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 same parts, are themselves equal and
parallel b; therefore EB, CH are both equal and parallel, and b. 33. 1.
EBCH is a parallelogram; and it is equal to ABCD, because it c. 35. 1.
is upon the same base BC, and between the fame parallels BC,
AD. for the like reason the parallelogram EFGH is equal to the
fame EBCH. therefore also the parallelogram ABCD is equal to
EFGH. Wherefore parallelograms, &c. Q. E. D.

T

PROP. XXXVII. THEOR.

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

Let the triangles ABC, DBC be upon the fame base BC and

figures EBCA, DBCF is a parallelogram; and EBCA is equal to b. 35. 1.

DBCF, because they are upon the fame base BC, and between the

fame parallels BC, EF; and the triangle ABC is the half of the pa

C

Book I. 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 d. 7. Ax. equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

C. 34. I.

T

PROP. XXXVIII. THEOR.

RIANGLES upon equal bases, and between the fame 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.

a. 31. 8.

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

C. 34. I.

lels BF, GH; 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 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.

:

PROP. XXXIX. THEOR.

EQUAL triangles upon the fame bafe, and upon the

fame fide of it, are between the fame parallels.

Let the equal triangles ABC, DBC be upon the fame base BC, and upon the fame fide 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 a AE parallel to BC, and join EC. the triangle ABC is equal to the triangle EBC, because it is upon the same base BC, Book I.

that no other line but AD is parallel to BC; AD is therefore pa

rallel to it. Wherefore equal triangles upon, &c.

E. D.

PROP. XL. THEOR.

EQUAL triangles upon equal bafes, and towards the

parts, are between the same parallels.

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

and join GF. the triangle ABC is equal to the triangle B

2. 31. 1.

GEF, because they are upon equal bases BC, EF, and between the fame parallels BF, AG. but the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the tri

b. 38. 1.

angle GEF, the greater to the less, which is impossible. therefore

AG is not parallel to BF. and in the fame manner it can be demonftrated that there is no other parallel to it but AD, AD is therefore parallel to BF. Wherefore equal triangles, &c. Q. E. D.

IF

PROP. XLI. THEOR.

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

Book I.

a. 37. 1.

b. 34. 1.

a. 10. 1.

b. 23. 1. C. 31. I.

d. 38. 1.

Let the parallelogram ABCD and the triangle EBC be upon the fame base BC, and between the fame parallels BC, AE; the

parallelogram ABCD is double of the A

triangle EBC.

DE

Join AC; then the triangle ABC is equal a to the triangle EBC, because they are upon the same base BC, and between the fame parallels BC, AE. but the parallelogram ABCD is double b of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double of the triangle EBC. therefore if a parallelogram, &c. Q. E. D.

O describe

B.

PROP. XLII. PRO B.

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. Bisect BC in E, join AE, and at the point E in the straight line EC make the angle CEF equal to D; and thro' A draw & AG

parallel to EC, and thro' C draw c

CG parallel to EF. therefore
FECG is a parallelogram. and be-
cause BE is equal to EC, the tri-
angle ABE is likewise equal d to
the triangle AEC, since they are
upon equal bases BE, EC and be-
tween the fame parallels BC, AG;

AF

4

G

D

B

EC

therefore the triangle ABC is dou

ble of the triangle AFC. and the parallelogram FECG is likewise double of the triangle AEC, because it is upon the fame 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 Lwherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of Book I. its angles CEF equal to the given angle D. Which was to be done.

PROP. XLIII. THEOR.

HE complements of the parallelograms which are

T about the diameter of any parallelogram, are

equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC,

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal a to the triangle ADC. and because EKHA 2. 34. 1. is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK. by the same reason, the triangle KGC is equal to the triangle KFC. then because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK together with the triangle KGC is equal to the triangle AHK together with the triangle KFC. but the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore the complements, &c. Q. E. D.

T

PROP. XLIV. PROB.

10 a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let AB be the given straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D.