Sidebilder
PDF
ePub
[merged small][merged small][merged small][merged small][ocr errors]

PROP. XXVII. THEOR.

IF a straight line falling upon two other straight lines make the alternate angles equal to one another, these two straight lines shall be parallel.

Let the straight line EF, which falls upon the two straight Fines AB, CD make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

[ocr errors]

E

B

pro

For, if it be not parallel, AB and CD being produced shallmeet either towards B, D, or towards A, C ; let them be duced and meet towards B, D,in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater a than the interior and opposite angle EFG; but it is also equal to it, which is impossible; A therefore AB and CD béing produced do not meet towards B, D. In like manner it may be demonstrated that they do not meet towards A, C; but those

F

D.

straight lines which meet neither way, though produced ever so far, are parallel to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXVIII. THEOR.

IF a straight line falling upon two other straight lines make the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines shall be parallel to one another.

A

-Let the straight line EF, which E
falls upon the two straight lines
AB, CD, make the exterior angle
EGB equal to the interior and op-
posite angle GHD upon the same
side; or make the interior angles
on the same side BGH, GHD to C
gether equal to two right angles;
AB is parallel to CD.

Because the angle EGB is equal
to the angle GHD, and the angle

[blocks in formation]

Book I.

EGB equal to the angle AGH, the angle AGH is equal to the angle GHD; and they are alternate angles; therefore AB is parallel to CD. Again, because the angles BGH, GHD are a 15. 1, equals to two right angles; and that AGH, BGH, are also b 27. 1. equal to two right angles; the angles AGH, BGH are equal G c By byp. to the angles BGH, GHD: take away the common angle BGH; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q, ED.

[merged small][ocr errors]

d 13. 1.

notes on this

IF a straight line fall upon two parallel straight See the lines, it makes the alternate angles equal to one ano- proposition. ther; and the exterior angle equal to the interior and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles, AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same

side GHD; and the two interior angles BGH, GHD upon A the same side are together equal to two right angles.

For if AGH be not equal to GHD, one of them must be C greater than the other; let AGH be the greater; and be

cause the angle AGH is great

E.

H

B

-AD

er than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles AGH, BGH are equal a to two a 13. 1. right angles; therefore the angles BGH, GHD are less than two right angles; but those straight lines which, with another straight line falling upon them, make the interior angles on * 12 ax. the same side less than two right angles, do meet together if See the continually produced; therefore the straight lines AB, CD, if notes on produced far enough, shall meet; but they never meet, since this propothey are parallel by the hypothesis; therefore the angle AGH is not unequal to the angle GHD, that is, it is equal to it; but b. 15. 1 the angle AGH is equal to the angle EGB; therefore like

sition.

Book I

13. 1.

a 29. 1.

b 27.1.

23. 1.

b27.1.

C

wise EGB is equal to GHD; add to each of those 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 also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D..

PROP. XXX. THEOR.

STRAIGHT lines which are parallel to the same straight line are parallel to one another.

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

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

a

lines AB, EF, the angle AGH is
equal to the angle GHF. Again,
because the straight line GK cuts
the parallel straight lines EF, CD,
the angle GHF is equal to the
angle GKD; and it was shown that
the angle AGK is equal to the an-
gle GHF; therefore also AGK is
equal to GKD; and they are alter-
nate angles; therefore AB is paral-

A

E

C

ور

B

H

Κ

lelb to CD, Wherefore straight lines, &c. Q. E. D.

PROP. XXXI. PROB.

TO draw a straight line through a given point pa

rallel to a given straight line.

Let A be the given point, and BC the given straight line;

it is required to draw a straight

line through the point A, parallel E

[blocks in formation]
[blocks in formation]

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

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

B

A

C

E31. 1.

a

Db 29. 1.

Through the point C draw CE parallel to the straight line AB; and because AB is parallel to CE and AC meets them, the alternate angles BAC, ACE are equalb. Again; because AB is parallel to CE, and BD falls upon them, the exterior 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, BAC, ACB; but the angles ACD, ACB are equal to two right angles: therefore also the angles c 13. 1. 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 to twice as many right angles as the figure has sides.

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding

E

D

A

B

Book I. 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 angies 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, together with four right angles, Therefore all the angics of the figure, together with four right angles are equal to twice as many right angles as the figure has sides.

a 2 Cor. 15. 1.

b 13. 1.

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 sides of the figure; D that is, by the foregoing corollary, they are equal to all the interior angles of the figure, to

A

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

PROP. XXXIII. THEOR..

a 29.1.

b 4. Ì.

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- A 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 them,

C

B

D

the alternate angles ABC, BCD are equala; 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 triangle ABC to the triangle BCD, and the other angles to the other angles, each to each, to which the equal sides are opposite: therefore the

« ForrigeFortsett »