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 draw BG parallel a to CA, and through F draw

FH parallel to

G

H

с

b xxxvi. 1.

с

and the triangle ABC is the half of the parallelogram GBCA, xxxiv. 1. because the diameter AB bisects it;

с

and the triangle DEF is the half of the parallelogram DEFH,

because the diameter DF bisects it:

But the halves of equal things are equal;

therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

d 7 Ax,

BOOK I.

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,

* xxxi. 1. through the point A drawa AE parallel to BC,

therefore also the triangle BDC is equal to the triangle 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 therefore parallel to it.

Wherefore equal triangles upon, &c. Q. E. D.

BOOK I.

* xxxi. 1.

b xxxviii. 1.

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 line BF, and towards the same

b

D

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.

E

BOOK I.

PROP. XLI. THEOR.

If a parallelogram and triangle be upon the same base, and between the same parallels:

be

The parallelogram shall be double of the triangle.

Let the parallelogram ABCD and the triangle EBC
upon the same base BC,

and between the same parallels BC, AE;

the parallelogram ABCD is double of
the triangle EBC.

Join AC;

a

* xxxvii. 1. then the triangle ABC is equal to

the triangle EBC,

A

D E

because they are upon the same base BC,

and between the same parallels BC, AE.

b

B

xxxiv. 1. But the parallelogram ABCD is double 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.

BOOK I.

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.

therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG

is likewise double of the triangle AEC,

because 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 parallelogram FECG equal to a given triangle ABC,

. x. 1.

b xxiii. 1.

• xxxi. 1.

d xxxviii. 1.

. xli. 1.