Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure shall be equal to one another, and the diameter BC shall bisect it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another. [I. 29.

B

And because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another. [I. 29. Therefore the two triangles ABC, BCD have two angles ABC, BCA in the one, equal to two angles DCB, CBĎ in the other, each to each, and one side BC is common to the two triangles, which is adjacent to their equal angles;

therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, namely, the side AB equal to the side CD, and the side AC equal to the side BD, and the angle BAC equal to the angle CDB.

[I. 26.

And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB,

the whole angle ABD is equal to the whole angle ACD. [Ax. 2. And the angle BAC has been shewn to be equal to the angle CDB.

Therefore the opposite sides and angles of a parallelogram are equal to one another.

Also the diameter bisects the parallelogram.

For AB being equal to CD, and BC common,

the two sides AB, BC are equal to the two sides DC, CB each to each;

and the angle ABC has been shewn to be equal to the angle BCD;

therefore the triangle ABCis equal to the triangle BCD, [I. 4. and the diameter BC divides the parallelogram ACDB into two equal parts.

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

Parallelograms on 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.

If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated at the same point D, it is plain that each of the parallelograms is double of the triangle BDC;

A

B

and they are therefore equal to one another.

D

F

[I. 34.

[Axiom 6.

But if the sides AD, EF, opposite to the base BC

of the parallelo

grams ABCD, EBCF be not terminated

at

the same point,

then, because

ABCD is a par

B

allelogram AD is equal to BC;

for the same reason EF is equal to BC;

therefore AD is equal to EF;

[I. 34.

[Axiom 1.

therefore the whole, or the remainder, AE is equal to the

whole, or the remainder, DF.

And AB is equal to DC;

[Axioms 2, 3.

[I. 34.

therefore the two sides EA, AB are equal to the two sides FD, DC each to each;

and the exterior angle FDC is equal to the interior and opposite angle EAB;

[I. 29.

therefore the triangle EAB is equal to the triangle FDC.

[I. 4.

Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and the remainders are equal; [Axiom 3. that is, the parallelogram ABCD is equal to the parallelogram EBCF.

Wherefore, parallelograms on the same base &c. Q.E.D.

PROPOSITION 36. THEOREM.

Parallelograms on 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.

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.

Therefore BE, CH are both equal and parallel.
Therefore EBCH is a parallelogram.

[I. 33.

[Definition. And it is equal to ABCD, because they are on the same base BC, and between the same parallels BC, AH. [I. 35.

For the same reason the parallelogram EFGH is equal to the same EBCH.

Therefore the parallelogram ABCD is equal to the parallelogram EFGH. [Axiom 1.

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

PROPOSITION 37. THEOREM.

Triangles on the same base, and between the same par› allels, are equal.

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.

Produce AD both ways to the points E, F; [Post. 2.

B

through B draw BE parallel to CA, and through C draw CF parallel to BD.

[I. 31.

Then each of the figures EBCA, DBCF is a parallelogram; [Definition and EBCA is equal to DBCF, 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, because the diameter AB bisects the parallelogram; [I. 34. and the triangle DBC is half of the parallelogram DBCF, because the diameter DC bisects the parallelogram. [I. 34. But the halves of equal things are equal. [Axiom 7. Therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles &c. Q.E.D.

Triangles on equal bases, and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be on equal bases BC, EF, and between the same parallels BF, AD: the triangle ABC shall be equal to the triangle DEF.

Produce AD both

ways to the points

D

G, H;

figures GBCA, DEFH is a parallelogram.

[Definition. 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 GBCA, because the diameter AB bisects the parallelogram; [I. 34. and the triangle DEF is half of the parallelogram DEFH, because the diameter DF bisects the parallelogram. But the halves of equal things are equal.

[Axiom 7. Therefore the triangle ABC is equal to the triangle DEF. Wherefore, triangles &c. Q.E.D.

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

Let the equal triangles ABC, DBC be on the same base BC, and on the same side of it: they shall be between the same parallels.

Then the triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE.

[I. 37.

But the triangle ABC is equal to the triangle DBC. [Hyp. Therefore also the triangle DBC is equal to the triangle EBC,

the greater to the less; which is impossible.

Therefore AE is not parallel to BC.

[Axiom 1.

In the same manner it can be shewn, that no other straight line through A but AD is parallel to BC; therefore AD is parallel to BC.

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

PROPOSITION 40. THEOREM.

Equal triangles, on equal bases, in the same straight line, and on the same side of it, are between the same parallels.

Let the equal triangles ABC, DEF be on equal bases BC, EF, in the same straight line BF, and on the same side of it: they shall be between the same parallels. Join AD.