Book I. EAF is drawn through the given point A parallel to the given

straight line BC. Which was to be done.

PROP. XXXII. THEOR.

If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles ; and the three interior angles of every triangle are equal to two right angles.

Let ABC be a triangle, and let one of its sides BC be produced to D; the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC, and the three interior angles of the triangle, viz. ABC, BCA, CAB are together equal to two right angles.

Through the point C a 31. 1. draw CE parallel a to the straight line AB; and be

A cause AB is parallel to CE

E and AC meets them, the al

ternate angles BAC, ACE b 29. 1. are equal. Again, because

AB is parallel to CE, and
BD falls upon them, the ex- B

с D
terior angle ECD is equal
to the interior and opposite angle ABC; but the angle ACE
was shown to be equal to the angle BAC; therefore the whole
exterior angle ACD is equal to the two interior and opposite
angles CAB, ABC; to these equals add the angle ACB,
and the angles ACD, ACB are equal to the three angles
CBA, BAČ, ACB; but the angles A CD, ACB, are equal
to two right angles; therefore also the angles CBA, BAC,
ACB are equal to two right angles. Wherefore, “ if a side
of a triangle,” &c. Q. E. D.

Cor. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal

D to twice as many right angles as

E
the figure has sides.

For any rectilineal figure
ABCDE can be divided into as

F
many triangles as the figure has
sides, by drawing straight lines
from a point F within the figure A B
to each of its angles. And, by

с

Book I.

15. 1.

the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure, and the same angles are equal to the angles of the figure together with the angles at the point F, which is the common vertex of the triangles; that is, a together with four right angles. Therefore, “all the a 2. Cor. “ angles of the figure together with four right angles, are equal “ to twice as many right angles as the figure has sides.”

Cor. 2. All the exterior angles of any rectilineal figure, are together equal to four right angles.

Because every interior angle ABC, with its adjacent exterior ABD, is equal to two right

b 13. 1. angles; therefore all the interior, together with all the exterior angles of the figure, are equal to A twice as many right angles as there are sides of the figure ; that is, by the foregoing corollary, they are equal to all the interior angles of the figure to- D B gether with four right angles ; therefore, “ all the exterior angles are equal to four right angles.”

PROP. XXXIII. THEOR.

А

The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and pa

B rallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel.

Join BC; and because AB is parallel to CD, and BC meets

С

D them, the alternate angles ABC, BCD are equal; a and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the base AC is equal to the base BD, and the tri- b 4. 1. angle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are

Book I. opposite : therefore the angle ACB is equal to the angle CBD;

and because the straight line BC meets the two straight lines

AC, BD, and makes the alternate angles ACB, CBD equal c 27, 1. to one another, AC is parallel to BD; and it was shown to

be equal to it. Therefore, “ straight lines,” &c. Q. E. D.

PROP. XXXIV. THEOR.

The opposite sides and angles of a parallelogram are equal to one another, and its diameter bisects it, that is, divides it into two equal parts.

N. B. A parallelogran is a four sided figure, of which the opposite sides are parallel; and the diameter is the straight line joining two of its opposite angles.

a 29. 1.

Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it. Because AB is parallel to

A

B
CD, and BC meets them, the
alternate angles ABC, BCD,
are equal a to one another; and
because AC is parallel to BD,
and BC meets them, the alter-
nate angles ACB, CBD are

с

D equal a to one another: wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and the side BC which is adjacent to these equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, 6 viz. the side AB to the side CD, and AC to BD, and the angle BAC equal to the angle BDC: 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: And the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of a parallelogram are equal to one another: also its diameter bisects it; for AB being 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 triangle ABC is Book I. equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.

c 4. 1.

PROP. XXXV. THEOR.

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

F

Let the parallelograms ABCD, EBCF be upon the same see the base BC, and between the same parallels AF, BC; the paral- 2d and 3d

figures. lelogram ABCD is equal to the parallelogram EBCF.

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

a 31. 1. triangle BDC; and they are therefore equal to one another.

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

B C ABCD, EBCF, be not terminated in the same point; then because ABCD is a parallelogram, AD is equal a to BC; for the same reason EF is equal to BC; wherefore AD is equal to' b 1. Ax. EF; and DE is common; therefore the whole, or the remainder AE is equal to the whole, or the remainder DF; AB also is c 2. or 3. equal to DC; and the two EA, AB are therefore equal to the

A D E F A E D F

Ax.

two FD, DC, each to each; and the exterior angle FDC is equald to the interior EAB, therefore the base EB is equal to d 29. 1. the base FC, and the triangle EAB equale to the triangle e 4. 1. FDC: take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders therefore are equal,' that is, the parallelogram f 3. Ax. ABCD is equal to the parallelogram EBCF. Therefore, “parallelograms upon the same base,” &c. Q. E. D.

Book I.

PROP. XXXVI. THEOR.

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

Let ABCD, EFGH A D E H
be parallelograms upon
equal bases BC, FG, and
between the same paral-
lels AH, BG; the paral-
lelogram ABCD is equal
to EFGH.
Join BE, CH; and B

C F

G because BC is equal to a 34. 1. FG, and FG to a EH, BC is equal to EH; and they are paral

lels, and joined towards the same parts by the straight lines BE, CH: but straight lines which join equal and parallel

straight lines towards the same parts, are themselves equal b 33. 1. and parallel; therefore EB, CH are both equal and parallel, c 35. 1. and EBCH is a parallelogram; and it is equal to ABCD, because it is

upon

the base BC, and between the same parallels BC, AH: For the like reason, the parallelogram EFGH is equal to the same EBCH: Therefore also the parallelogram ABCD is equal to EFGH. Wherefore, “ pa“ rallelograms,” &c. Q. E. D.

same

PROP. XXXVII. THEOR.

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

Let the triangles ABC, DBC be upon the same base BC, and between the

same parallels AD, BC: The triangle

E AD F ABC is equal to the triangle DBC.

Produce AD both ways

to the points E, F, and a 31. 1. through B draw a BE parallel to CA; and through

B

C
C draw CF parallel to BD:

Therefore each of the figures EBCA, DBCF is a paralb 35. 1. lelogram; and EBCA is equal to DBCF, because they are

upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram