the opposite sides and angles of parallelograms are equal to one another, also the diameter bisects it. For because AB is equal to CD, and BC common, the two AB, BC, are equal to the two DC, CB, each to each; and the angle ABC is equal to the angle BCD. Therefore the base AC is equal to the base DB, and the triangle ABC is equal to the triangle BCD. Therefore the diameter BC bisects the parallelogram ABCD. Q.E.D.

Deductions.

1. The diameters of a parallelogram bisect each other. 2. If the corresponding diameters of two equiangular parallelograms be equal to one another, also a side of the one equal to the corresponding side of the other, then shall the other opposite sides of the one be equal to the other opposite sides of the other.

3. If in the sides of a square, at equal distances from the four angles, four other points be taken, one in each side, the figure contained by the right lines which join them shall also be a square.

PROPOSITION XXXV.

THEOREM.

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

A D E F

B

Let ABCD, EBCF, be parallelograms, placed upon the same base BC, and between the same parallels AF, BC. The parallelogram ABCD is equal to the parallelogram EBCF. For because ABCD is a parallelogram, AD is equal to BC. For the same reason EF is equal to BC. And therefore AD will be equal to EFb and DE common. Therefore the whole AE is equal to the whole DF. But AB is equal to DC. Therefore the two EA, AB, are equal to the two FD, DC, each to each, and the angle EDC is equal to the angle EAB, the exterior to the interior; therefore the base EB is equal to the base Fc, and the triangle EAB is equal to the triangle FDC. Take away DGE, which is common. Therefore the remaining trapezium ABGD is equal to the remaining trapezium EGCF.f Add the triangle GBC, which is common.

Therefore the whole parallelogram ABCD will be equal to the whole parallelogram EBCF. Therefore parallelograms placed upon the same base, &c. q. E. d.

PROPOSITION XXXVI.

THEOREM.

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

A

DE H

• Hyp.

B C F G

Let ABCD, EFGH, be parallelograms constituted upon equal bases BC, FG, and between the same parallels AH, BG. The parallelogram ABCD is equal to the parallelogram EFGH. For join BE, CH; and because BC is equal to FG,a and FG equal to EH, BC will also be equal to EH. Therefore EB, CH, are both equal and parallel. But those lines are parallel which join the extremities of equal and parallel right lines towards the same parts. Therefore EB, CH, are equal and parallel; wherefore EBCH is a parallelogram, and it is equal to the parallelogram ABCD, for it is placed upon the same base BC, and between the same parallels BC, AD. For the same reason the parallelogram EFGH is equal to the parallelogram EBCH, for 35. 1. it has the same base EH, and is constituted between the same parallels EH, BG. Therefore the parallelogram ABCD will be equal to the parallelogram EFGH. There

fore parallelograms, &c.

Q. E. D.

Deductions.

1. If the sides of a parallelogram be bisected, the lines joining the opposite points of section will divide the parallelogram into four equal parallelograms.

2. If a trapezium have two of its sides parallel to one another, and equal to two sides of another trapezium, which are parallel to one another; also if the perpendicular distance of the one be equal to the perpendicular distance of the other, then shall the two trapeziums be equal to one another..

31. 1.

35. 1.

• 34. 1.

a 31. 1.

b 36. 1.

€ 34. 1.

d Ax. 7.

PROPOSITION XXXVII.

THEOREM.

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

a

E

AD

F

W

B

Let the triangles ABC, DBC, be constituted upon the same base BC, and between the same parallels AD, BC. The triangle ABC is equal to the triangle DBC. Produce AD both ways to the points E, F ; and through в draw вE, parallel to CA, and through c draw CF, parallel to BD. Therefore each of them EBCA, DBCF, is a parallelogram, and the parallelogram EBCA is equal to the parallelogram DBCF. For they are upon the same base BC, and between the same parallels BC, EF. And the triangle ABC is half of the parallelogram EBCA, because the diameter AB bisects it; and the triangle DBC is half of the parallelogram DBCF, for the diameter DC bisects it. But the halves of equal things are equal. Therefore the triangle ABC is equal to the triangle DBC. Therefore triangles, &c. Q. E. D.

PROPOSITION XXXVIII.

THEOREM.

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

A

D H

B CE

Let the triangles ABC, DCE, be constituted upon equal bases BC, CE, and G between the same parallels BE, AD. The triangle ABC is equal to the triangle DCE. For produce AD both ways to the points G, H. Through в draw BG parallel to CA; also through E draw EH parallel to DC. Therefore each of the figures GBCA, DCEH, is a parallelogram. And the parallelogram GBCA is equal to the parallelogram DCEH, because they are upon equal bases BC, CE, and between the same parallels BE, GH. But the triangle ABC is half of the parallelogram GBCA, for the diameter AB bisects it; and the triangle DCE is half of the parallelogram DCEH, for the diameter DE bisects it. But the halves of equal things are equal: therefore

b

the triangle ABC is equal to the triangle DCE. Therefore triangles, &c. Q. E. D.

Deductions.

1. A right line drawn from the vertex of a triangle bisecting the base, divides the triangle into two equal triangles.

2. If two opposite sides of a trapezium be parallel to one another, the line joining their bisections bisects the trapezium.

PROPOSITION XXXIX.

THEOREM.

Equal triangles placed upon the same base, and towards the same parts, are also between the same parallels.

EC.

A

B

E

D

a 31. 1.

Let the equal triangles ABC, DBC, be constituted upon the same base BC, and towards the same parts they are between the same parallels. For draw AD; AD is parallel to BC. For if it is not parallel, draw through the point A the right line AE, parallel to BC, and join Therefore the triangle ABC is equal to the triangle EBC, because it is upon the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal to the 37. 1. triangle DBC. 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. In like manner we show that none other, except AD, is parallel to BC. Whence AD is parallel to BC. Therefore equal triangles, &c. Q. E. D.

PROPOSITION XL.

THEOREM.

D

Equal triangles constituted upon equal bases, and towards the same parts, are also between the same parallels. Let the equal triangles ABC, CDE, be constituted on the equal bases BC, CE. A They are between the same parallels. Draw AD; AD is parallel to BE. For if it is not, through A draw AF parallel to BE, and join DE. Therefore the triangle B ABC is equal to the triangle CEF, because

b

M...

E

a 31. 1.

b 38.1.

. 37. 1.

34. 1.

10. 1.

they are constituted between the same parallels BE, AF,
and
upon equal bases. But the triangle ABC is equal
to the triangle DCE. Therefore also the triangle DCE
will be equal to the triangle FCE, the greater to the less,
which is impossible. Therefore AF is not parallel to BE.
In like manner we may show that no other line drawn
through A is parallel to BE except AD. Therefore AD
will be parallel to BE. Therefore equal triangles, &c.

Q. E. D.

PROPOSITION XLI.

THEOREM.*

If a parallelogram and a triangle have the same base, and are between the same parallels, the parallelogram will be double of the triangle.

A D E

B

For let ABCD be a parallelogram, and EBC a triangle; let them have the same base BC, and between the same parallels BC, AE. The parallelogram ABCD, is double of the triangle EBC. For join AC. Therefore the triangle ABC is equal to the triangle EBC," for they are constituted upon the same base BC, and between the same parallels BC, AE. But the parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects it.b Wherefore it will also be double of the triangle EBC. If therefore a parallelogram and a triangle, &c. Q. E. d.

PROPOSITION XLII.

PROBLEM.

To make a parallelogram equal to a given triangle, and having one of its angles equal to a given rectilineal angle.

A F G

Let ABC be a given triangle, and D
a given rectilineal angle. It is re-
quired to make a parallelogram equal
to the triangle ABC, and having one B E
of its angles equal to the given recti-

lineal angle D. Bisect BC in E; and AE being joined
to the right line EC and to the point in it E, make the

*From this proposition is derived the rule for finding the area of a triangle, the base and altitude being given; for as the area of a parallelogram is the product of the base and altitude, it follows that the area of a triangle must be half that product.