line EAF is drawn through the given point A parallel to the Book I. given ftraight line BC. Which was to be done.

PROP. XXXII. THEOR.

Iangle is equal to the two interior and oppofite
F a fide of any triangle be produced, the exterior

angles; and the three interior angles of every tri-
angle are equal to two right angles.

Let ABC be a triangle, and let one of its fides BC be produced to D; the exterior angle ACD is equal to the two interior and oppofite 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 draw CE parallel a to the ftraight line AB; and because AB is parallel to CE and AC meets them, the alternate angles BAC, B

C

a 31. I.

E

D

ACE are equalb. Again, because AB is parallel to CE, and b 29. 1. BD falls upon them, the exterior angle ECD is equal to the interior and oppofite 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 oppofite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB; but the angles ACD, ACB are equal to two c 13 I. right angles; therefore alfo the angles CBA, BAČ, ACB are equal to two right angles. Wherefore, if a fide of a triangle, &c. Q. E. D.

COR. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has fides.

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has

E

D

C

F

C

fides, by drawing ftraight lines

from a point F within the figure

A

B

all

to each of its angles. And, by the preceding propofition,

D

Book I.

all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are fides of the figure; and the fame angles are equal to the angles of the figure, together with the angles at the point F, which is a 2. Cor. the common vertex of the triangles: that is a, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides.

15. I.

b 13. 1.

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

Becaufe every in

terior angle ABC,
with its adjacent ex-
terior ABD, is equal b
to two right angles;
therefore all the inte-
rior, together with all
the exterior angles of

the figure, are equal

to twice as many right D

angles as there are

fides of the figure;

that is by the forego

ing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles.

THE

'HE ftraight lines which join the extremities of two equal and parallel ftraight lines, towards the fame parts, are alfo themselves equal and parallel.

Let AB, CD be equal and A parallel ftraight lines, and joined towards the fame parts by the ftraight lines AC, BD; AC, BD are alfo equal and parallel.

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

D

a 29. 1. them, the alternate angles ABC, BCD are equala; and be

↑ caufe

1

cause AB is equal to CD, and BC common to the two tri- Book I.
angles ABC, DCB, the two fides AB, BC are equal to the
two DC, CB; and the angle ABC is equal to the angle BCD;
therefore the base AC is equalb to the bafe BD, and the tri- b 4. I.
angle ABC to the triangle BCD, and the other angles to the
other angles b, each to each, to which the equal fides are op-
pofite; 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
to one another, AC is parallel to BD; and it was fhewn to c 27. I.
be equal to it. Therefore, straight lines, &c. Q. E. D.

'HE oppofite fides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them in two equal parts.

N. B. A parallelogram is a four fided figure, of which the oppofite fides are parallel; and the diameter is the ftraight line joining two of its oppofite angles.

Let ACDB be a parallelogram, of which BC is a diameter; the oppofite fides and angles of the figure are equal to one another; and the diameter BC bifects it.

Because AB is parallel to CD, A

and BC meets them, the alternate angles ABC, BCD are equal a to one another; and becaufe AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal a

B

to one another; wherefore the two triangles ABC, CBD havetwo angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one fide BC common to the two triangles, which is adjacent to their equal

D 2

angles;

a 29. 1.

Book I.

b 26. I.

C 4. I.

See the 2d'

and 3d figures.

angles; therefore their other fides fhall be equal, each to
each, and the third angle of the one to the third angle of the
other b, viz. the fide AB to the fide 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
fhown to be equal to the angle BDC; therefore the oppofite
fides and angles of parallelograms are equal to one another:
also, their diameter bifects them; 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 equal to the triangle
BCD, and the diameter BČ divides the parallelogram ACDB
into two equal parts. Q. E. D.
Q.E.

PARAL

PROP. XXXV. THEOR.

ARALLELOGRAMS upon the fame base and between the fame parallels, are equal to one a

nother.

Let the parallelograms ABCD, EBCF be upon the fame bafe BC, and between the fame parallels AF, BC; the parallelogram ABCD fhall be equal to the parallelogram EBCF.

If the fides AD, DF of the A parallelograms ABCD, DBCF oppofite to the base BC be terminated in the fame point D; it is plain that each of the paa 34. 1. rallelograms is double a of the triangle BDC; and they are therefore equal to one another.

B

D

F

But, if the fides AD, EF, oppofite to the bafe BC of the parallelograms ABCD, EBCF, be not terminated in the fame point; then, because ABCD is a parallelogram, AD is equal a to BC; for the fame reason EF is equal to BC; wherefore AD is equal to EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or

the

the remainder DF; AB alfo is equal to DC; and the two

Book I.

EA, AB are therefore equal to the two FD, DC, each to each; and the exterior angle FDC is equal d to the interior d 19. 1. EAB, therefore the base EB is equal to the base FC, and the triangle EAB equal e to the triangle FDC; take the tri- e 4. 1. angle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB; the remainders therefore are equal, that is, the parallelogram ABCD is equal to the f 3. Ax, parallelogram EBCF. Therefore, parallelograms upon the fame bafe, &c. Q. E. D.

ARALLELOGRAMS upon equal bafes, and be-
tween the fame parallels, are equal to one ano-

PAR

ther.

Let ABCD, EFGH A be parallelograms upon equal bafes BC, FG, and between the fame parallels AH, BG; the parallelogram APCD is equal to EFGH.

Join BE, CH; and B

because BC is equal to FG, and FG to a EH, BC is equal to a 34. I. EH; and they are parallels, and joined towards the fame parts by the ftraight lines BE, CH: But ftraight lines which join equal and parallel ftraight lines towards the fame parts, are themselves equal and parallel; therefore EB, CH are b 33. 1. both equal and parallel, and EBCH is a parallelogram; and

C

it is equal to ABCD, because it is upon the fame base BC, c 35. 1. and

D 3