Book I. 15

C 13. I.

of thefe the angle EGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal e to two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a straight, &c. Q. E. D. PROP. XXX. THEOR.

ST

TRAIGHT lines which are parallel to the fame ftraight line are parallel to one another.

Let AB, CD be each of them parallel to EF; AB is also parallel to CD.

Let the ftraight line GHK cut AB, EF, CD; and because GHK cuts the parallel ftraight

lines AB, EF, the angle AGH is

E

a 29. I. equal a to the angle GHF. A-
gain, because the ftraight line A-
GK cuts the parallel ftraight lines
EF, CD, the angle GHF, is equal
a to the angle GKD; and it was
fhewn that the angle AGK is e-
qual to the angle GHF; there-
fore alfo AGK is equal to GKD;
and they are alternate angles;

C

b 27. 1. therefore AB is parallel to CD. Wherefore ftraight lines,

&c. Q. E. D.

T

PROP. XXXI. PROB.

draw a fraight line through a given point parallel to a given ftraight line.

A F

Let A be the given point, and BC the given ftraight line;
it is required to draw a flraight line
through the point A, parallel to the
ftraight line BC.

In BC take any point D, and join
AD; and at the point A, in the

E

a 23. 1. ftraight line AD make a the angle B

DAE equal to the angle ADC; and

produce the traight line EA to F.

D

C

Because the ftraight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to

b 27. I. one another, EF is parallel b to BC. Therefore the straight line

EAF

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

PROP. XXXII. THE OR.

F a fide 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 fides BC be produced to D; the exterior angle ACD is equal to the two inte rior 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 straight line AB; and because AB is parallel to CE and AC meets them, the alternate angles BAC, ACE are equal b. Again, because AB is parallel to CE, and BD falls uponthem, the exterior angle ECD is equal to the interior and

oppofite angle ABC; but the angle ACE was fhewn 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 thefe equals 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 right angles; therefore alfo the c 13. I. angles CBA, BAC, ACB are equal to two right angles. Wherefore if a fide of a triangle, &c. Q. E. D.

Book I. the preceding propofition, 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 a 2. Cor. F, which is 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.

b13. I.

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
angles; therefore all the interi-
or, together with all the exterior
angles of the figure, are equal to
twice as many right angles as
there are fides of the figure;
that is, by the foregoing corol-D

lary, they are equal to all the
interior angles of the figure, to-

A

B

gether with four right angles; therefore all the exterior angles are equal to four right angles.

4 29. I.

b 4. I.

THE

HE ftraight lines which join the extremities of twa equal and parallel ftraight lines, towards the fame

parts, are alfo themselves equal and parallel.

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

B

D

Join BC; and because AB is parallel to CD, and BC meets 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 fides AB, BC are equal to the two DC, CB; and the angle ABC is equal to the angle BCD; therefore the bafe AC is equal b to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles b, each to each, to which the equal fides are oppofite; therefore the angle

angle ACB is equal to the angle CBD; and because the ftraight Book I. line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BĎ; and it was fhewn to be equal to it. Therefore, c 27. x. ftraight lines, &c. Q. E. D.

TH

PROP. XXXIV. THEOR.

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

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

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

Because AB is parallel to CD, A

C

B

a 29. I.

D

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 alternate angles ACB, CBD are 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 one fide BC common to the two triangles, which is adjacent to their equal 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, viz. the fide b 26. F. 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 fhewn to be equal to the angle BDC; therefore the oppofite fides and angles of parallelograms are equal to one another; alfo, 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 iş

C4

equal

Book I. equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.

C 4. I.

See N.

See the 2d

gures.

PAR

PROP. XXXV. THEOR.

ARALELLOGRAMS upon the fame base and between the fame parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the fame bafe

and 3d fi- 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 parallelograms ABCD, DBCF oppofite to A

a 34. I.

b 1. Ax.

€ 2. or 3. Ax.

the bife BC be terminated in the same
point D; it is plain that each of the
parallelograms is double a of the tri-
angle BDC; and they are therefore
equal to one another.

D

B

C

F

But, if the fides AD, EF, oppofite to the base BC of the parallelograms ABCD. EBCF, be not terminated in the fame point; then, be-. cause ABCD is a parallelogram, AD is equal a to BC; for the fame reafon EF is equal to BC; wherefore AD is equal b to EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or the remainder DF; AB alfo is equal to DC; and the two EA, AB are therefore equal to D F

с

DE

FA E

B

B

C

d 19. 1. € 4. I.

f 3. Ax.

the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB, therefore the base EB is equal to the base FC, and the triangle EAB equal e to the triangle FDC; take the triangle FDC from the trapezium ABCF, and from the fame trapezium take the triangle EAB; the remainders therefore are equal f, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the fame bale, &c. · Q. E. D.

PROP.