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C 13. I.

a 29. I.

b 27. 1.

each of these the angle BGH; therefore the angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal to two right angles; therefore alfo BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D.

STR

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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 alfo 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

a

is equal to the angle GHF.
Again, because the fraight line A-
GK cuts the parallel ftraight lines
EF, CD, the angle GHF is equal E
to the angle GKD; and it was

a

fhewn that the angle AGK is e- C
qual to the angle GHF; there
fore alfo AGK is equal to GKD;
and they are alternate angles;

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therefore AB is parallel to CD. Wherefore ftraight lines, &c.

Q. E. D.

T

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O draw a ftraight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw a ftraight line

through the point A, parallel to the
ftraight line BC.

E

A

F

a 23. I.

b 27. J.

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

a

straight line AD make the angle BD
DAE equal to the angle ADC; and

produce the ftraight line EA to F.

C

Because the ftraight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, Ek is parallel to BC. Therefore the ftraight line

EAF

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

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IF a fide of any triangle be produced, the exterior angle is equal to the two interior and oppofite 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 pro. duced 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 to the ftraight line AB; and because AB is parallel to CE and AC meets them, the alternate angles BAC, ACE are equal b. Again, becaufe AB is parallel to CE, and BD falls upon them, the exterior angle ECD B is equal to the interior and

a 31. 1.

A

E

b 29. I.

D

oppofite angle ABC; but the angle ACE was fhown 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 allo the c 13. 1. angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a fide of a triangle, &c. Q. E. D.

с

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

For ABCDE can be divided into as many triangles as the figure has fides, by drawing ftraight lines from a point F within the figure

any rectilineal figure

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Book I. to each of its angles. And, by the preceding propofition, all w the angles of these triangles are equal to twice as many right

a 2. Cor. 15. I.

b 13. I.

a 29 I.

b 4. 1.

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 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.

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 interior,
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 corollary, D B
they are equal to all the inte

rior angles of the figure, toge

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

TH

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

parts, are also 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.

Join BC; and becaufe ABis pa-
rallel to CD, and BC meets them,
the alternate angles ABC, BCD

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B

D

are equal; 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 to the base BD, and the triangle ABC to the triangle BCD, and the other angles

b

to

to the other angles b, each to each, to which the equal fides are Book I. oppofite; therefore the angle ACB is equal to the angle CBD; and because the ftraight line BC meets the two ftraight lines b 4. I. AC, BD, and makes the alternate angles ACB, CBD equal to

one another, AC is parallel to BD; and it was fhown to be c 27.1. equal to it. Therefore ftraight lines, &c. Q. E. D.

TH

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HE opposite fides and angles of parallelograms are equal to one another, and the diameter bifects 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 Straight 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 an other; and the diameter BC bifects it.

Because AB is parallel to CD,

a

a

C

B

D

a 29. I.

and BC meets them, the alternate angles ABC, BCD are equal to one another; and becaufe AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal * 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 b, viz. b 26. I. 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 shown to be equal to the angle BDC; therefore the opposite fides and angles of parallelograms are e. qual 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

C 4

is

Bo I. is equal to the angle BCD; therefore the triangle ABC is é qual 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

P

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

Let the parallelograms ABCD, EBCF be upon the fame

and 3d fi bafe BC, and between the fame parallels AF, BC; the paralle logram ABCD fhall be equal to the parallelogram EBCF.

gures.

2. 34. I.

b 1. Ax.

c 2. or 3. Ax.

If the fides AD, DF of the pa-
rallelograms ABCD, DBCF oppofite A

to the base BC be terminated in the
fame point D; it is plain that each
of the parallelograms is double a of
the triangle BDC; and they are there-
fore equal to one another

D

C

F

But, if the fides AD, EF, oppofite B 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 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

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d 29.3.

e 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 bafe EB is equal to the bafe FC, and the triangle EAB equal 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 bafe, &c. Q. E. D.

PROP.

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