For AB being equal to CD, and BC common,

The two sides AB, BC are equal to the two sides CD and CB, each to each.

And the angle ABC has been shown to be equal to the angle BCD;

Therefore the triangle ABC is equal to the triangle BCD (I. 4),

And the diagonal BC divides the parallelogram ABCD into two equal parts.

Therefore, the opposite sides, &c. Q. E. D.

Proposition 35.-Theorem.

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

Let the parallelograms ABCD, EBCF be on the same base BC, and between the same parallels AF, BC;

The parallelogram ABCD shall be equal to the parallelogram EBCF.

D

CASE 1.-If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D, it is plain that each of the parallelograms is double of the triangle DBC (I. 34), and that they are therefore equal to one another (Ax. 6).

B

CASE 2.-But if the sides AD, EF, opposite to the base BC, of the parallelograms ABCD, EBCF, be not terminated in the same point, then- A

PROOF.

- Because

DE

For the same reason EF is equal to BC;

FA E D

B

Therefore AD is equal to EF (Ax. 1), and DE is common;
Therefore the whole, or the remainder, AE, is equal to the

whole, or the remainder, DF (Ax. 2, or 3),

And AB is equal to DC (I. 34).

Therefore the two EA, AB are equal to the two FD, DC, each to cach;

also
A ABC =

A BCD.

AD BC.

EF= BC.

..AE=DF.

Hence

A EAB=
A FDC.

And the exterior angle FDC is equal to the interior EAB (I. 29);

Therefore the base EB is equal to the base FC (I. 4), And the triangle EAB equal to the triangle FDC (I. 4). Take the triangle FDC from the trapezium ABCF, and from the same trapezium ABCF, take the triangle EAB, and the remainders are equal (Ax. 3)

That is, the parallelogram ABCD is equal to the parallelogram EBCF.

Therefore, parallelograms, &c. Q. E. D.

Proposition 36.-Theorem.

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

Let ABCD, EFGH be parallelograms on equal bases BC, FG, and between the same parallels AH, BG;

The parallelogram ABCD shall be equal to the parallelogram EFGH.

BC= EH,

B

and

BE CHI.
EBCH a

parallelo

gram,

A

D E

H

C

F

CONSTRUCTION.--Join BE, CH. PROOF. Because BC is equal to FG (Hyp.), and FG to EH (I. 34),

Therefore BC is equal to EH G (Ax. 1); and they are parallels, and joined towards the same parts by the straight lines BE, CH.

But straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves. equal and parallel (I. 33);

Therefore BE, CH are both equal and parallel;
Therefore EBCH is a parallelogram (Def. 35),

And it is equal to the parallelogram ABCD, because they equal'each are on the same base BC, and between the same parallels BC, AH (I. 35).

of the

given ones.

For the like reason, the parallelogram EFGH is equal to the same parallelogram EBCH;

Therefore the parallelogram ABCD is equal to the parallelogram EFGH (Ax. 1).

Therefore, parallelograms, &c. Q. E. D.

Proposition 37.-Theorem.

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

Let the triangles ABC, DBC be on the same base BC, and between the same parallels AD, BC;

The triangle ABC shall be equal to the triangle DBC. CONSTRUCTION.-Produce AD both ways, to the points E, F. Through B draw BE parallel to CA, and through C draw CF parallel to BD (I. 31).

PROOF. Then each of the E figures EBCA, DBCF, is a parallelogram (Def. 35), and they are equal to one another, because they are on the same base BC, and between the same parallels BC, EF (I. 35.);

And the triangle ABC is half of the parallelogram EBCA, and the because the diagonal AB bisects it (I. 34);

And the triangle DBC is half of the parallelogram DBCF, because the diagonal DC bisects it (I. 34).

But the halves of equal things are equal (Ax. 7);
Therefore the triangle ABC is equal to the triangle DBC.
Therefore, triangles, &c. Q. E. D.

Proposition 38.-Theorem.

Triangles upon equal bases, and between the same paralle's, are equal to one another.

Let the triangles ABC, DEF, bo on equal bases BC, EF, and between the same parallels BF, AD.

The triangle ABC shall be equal to the triangle DEF.
CONSTRUCTION.-Produce AD both ways to the points

G, H.

Through B draw BG parallel to CA, and through F draw FH parallel to ED (I. 31).

PROOF.-Then each of the figures GBCA, DEFH, is a

B

D

n

triangles are respec

tively half

' of these.

Figures
GBCA and
DEFH are
equal;

and the

parallelogram (Def. 35), and they are equal to one another, because they are on equal bases BC, EF, and between the same parallels BF, GH (I. 36);

And the triangle ABC is half of the parallelogram triangles GBCA, because the diagonal AB bisects it (I. 34);

are half of

these re

And the triangle DEF is half of the parallelogram spectively. DEFH, because the diagonal DF bisects it (I. 34).

AE parallel

to BC suppose.

Then

A DBC=

Δ

But the halves of equal things are equal (Ax. 7);
Therefore the triangle ABC is equal to the triangle DEF.
Therefore, triangles, &c. Q. E. D.

Proposition 39.-Theorem.

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

Let the equal triangles ABC, DBC be upon the same base
BC, and on the same side of it ;

They shall be between the same parallels.
CONSTRUCTION.—Join AD; AD shall be parallel to BC.

A

B

E

For if it is not, through A draw AE parallel to BC (I. 31), and join EC.

PROOF.-The triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the same c parallels BC, AE (I. 37).

But the triangle ABC is equal to the triangle DBC (Hyp.);
Therefore the triangle DBC is equal to the triangle EBC

A EBC, an (Ax. 1), the greater equal to the less, which is impossible;

absurdity.

Therefore AE is not parallel to BC.

In the same manner, it can be demonstrated that no line passing through A can be parallel to BC, except AD ; Therefore AD is parallel to BC.

Therefore, equal triangles, &c. Q. E. D.

Proposition 40.-Theorem.

Equal triangles upon the same side of equal bases, that are in the same straight line, are between the same parallels.

Let the equal triangles ABC, DEF, be upon the same side of equal bases BC, EF, in the same straight line BF.

The triangles ABC, DEF shall be between the same parallels.

CONSTRUCTION.-Join AD; AD shall be parallel to BF.

AG paral

For if it is not, through A draw AG parallel to BF (I. 31), lel to BF and join GF.

PROOF. The triangle ABC is equal to the triangle GEF, because they are upon equal bases BC, EF, and are between the same parallels BF, AG

(I. 38).

Therefore AG is not parallel to BF.

In the same manner, it can be demonstrated that no line, passing through A, can be parallel to BF, except AD; Therefore AD is parallel to BF.

Therefore, equal triangles, &c.

Proposition 41.-Theorem.

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

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

The parallelogram ABCD shall be double A of the triangle EBC.

CONSTRUCTION. -Join AC.

PROOF. The triangle ABC is equal to

the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE (I. 37).

D

E

suppose.

A DEF= A GEF, an absurdity.

A ABC =
A EBC.

But the parallelogram ABCD is double of the triangle And paralABC, because the diagonal AC bisects the parallelogram (I. 2 ABC.

34).

lelogram=