the third angle BAC to the third angle EDF. Therefore, if two triangles, &c. Q.E.D.

PROP. XXVII. THEOR.
If a straight line falling upon two other straight

lines makes the alternate angles equal to one
another, these two straight lines shall be pu-
rallel.

Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another: AB shall be parallel to CŨ.

For, if it be not parallel, AB and CD being produced will meet either towards B, D, or towards A, C: let them be produced and meet

towards B, D in the point G; therefore GEF is a * 16. 1. triangle, and its exterior angle AEF is greater*

than the interior and opposite angle EFG ; but it + Hyp. is also equals to it, which is

impossible; therefore AB A E B
and CD being produced do

G
not meet towards B,D. In CF
like manner it may be de-
monstrated, that they do not meet towards A, C:

but those straight lines which meet neither rway, *35 Def, though produced ever so far, are parallel * to

one another: therefore AB 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 makes the exterior angle equal to the in-
terior 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.
Let the straight line EF, which falls upon

the two straight lines AB, CD, make the exterior

A

G

*

angle EGB equal to the interior and opposite
angle GHD upon the same
side ; or make the interior

E angles on the same side BGH, GHD together equal

B to two right angles : AB

C

-D shall be parallel to CD.

H Because the angle EGB is equal t to the angle GHD,

+Нур and the angle EGB equal*, to the angle AGH, *15. 1. therefore the angle AGH is equal t to the angle +1 Ax. GHD: and they are the alternate angles; therefore AB is parallel*

to CD. Again, because the *27. 1. angles BGH, GHD are equal* to two right * Hyp. angles, and that AGH, BGH, are also equal * to *13. 1. two right angles; therefore the angles AGH, BGH are equalf to the angles BGH, GHD: f1 Ax Take away the common angle BGH; therefore the remaining angle AGH is equal † to the +3 Ax. remaining angle GHD: and they are alternate angles : therefore ABis parallelt to CD. Where- +27. 1. fore, if a straight line, &c. Q. E. D.

PROP. XXIX. THEOR. If a straight line fall upon two parallel straight See the

lines, it makes the alternate angles equal to one another; and the exterior angle equal to the position interior and opposite upon the sume side ; and Oct. Ed. 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, shall be equal to one another: and the exterior angle EGB shall be equal to the Interior and opposite, upon he same side, GHD; and the E two interior angles BGH, GHD upon the same side, A shall be together equal to two right angles.

H D For, if AGH be not equal

F

notes on

this pro

on

Oct. Ed.

to GHD, one of them must be greater than the other: let AGH be the greater: and because the angle AGH is greater than the angle GHD, add

to each of them the angle BGH; therefore the +4 Ax. angles AGH, BGH are greater than theangles

BGH, GHD: but the angles AGH, BGH, are *13. l. equal* to two right angles; therefore the angles

BGH, GAD, are less than two right angles : but those straight lines which, with another straight line falling upon them, make the in

terior angles on the same side less than two *12 Ax. right angles, will meet * together if continually Sereshen produced ; therefore the straight lines AB, CD), this pro- if produced far enough, will meet : but they position

never meet, since they are parallel by the hypothesis ; therefore the angle AGH is not un

equal to the angle GHD, that is, it is equal to *15. 1. it: but the angle AGH is equal * to the angle +1 Ax. EGB; therefore likewise EGB is equal t to

GHD: add to each of these the angle BGH; +2 Ax. therefore the angles EGB, BGH are equal t to

the angles BGH, GHD: but EGB, BGH are *13. 1. equal* to two right angles; therefore also BGH, *1 Ax. GHD are equal * to two right angles. Wherefore, if a straight line, &c. Q. E. D.

PROP. XXX. THEOR.
Struight lines which are parallel to the sume

straight line are parallel to each other.
Let AB, CD be each of them parallel to EF:
AB shall be parallel to CD.

Let the straight line GHK cut AB, EF, CD: and because GẮK cuts the parallel straight lines

AB, EF, the angle AGH is * 29. 1. equal * to the angle GHF. Again, because the straight

A

-B line GK cuts the parallel

Н straight lines EF, CD, the E

-F *28. 1. angle GHF is equal* to the C

D angle GKD:

and it was shown that the angle AGK

is equal to the angle GHF; therefore also AGK is equal * to GKD: and they are alternate +1 Ax. angles; therefore AB is parallel * to CD. * 27. 1. Wherefore, straight lines, &c. Q. E. D.

PROP. XXXI. PROB.
To draw a straight 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 straight
line through the point A, pa-
rallel to the straight line BC.

E A F In BC take any point D, and join AD; and at the point A, in the straight line AD,

make*

B 1) the angle DAE equal to the angle ADC; and produce the straight line EA to F: EF shall be parallel to BC.

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 * to BC. Therefore the straight *27. 1. line EAF is drawn through the given point A, parallel to the given straight line BC. Which was to be done.

C *23.1.

*

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 shall be equal to the two interior and opposite angles CAB, ABC: and the three interior angles of the triangle, viz. ABC, BCA, CAB, shall together be equal to two right angles.

:

Through the point C *31. 1. draw CE parallel* to the

E straight line AB: and because AB is parallel to CE, and AC meets them,

D the alternate angles BAC, * 29. 1. ACE, are equal.* Again, because AB is parallel

to CE, and BD falls upon them, the exterior + 29. 1. angle ECD is equal t to the interior and opposite

angle ABC : but the angle ACE was shown to

be equal to the angle BAČ; therefore the whole +2 Ax., exterior angle ACD is equal to the two interior

and opposite angles CAB, ABC: to each of these

equals add the angle ACB, and the angles ACD, +2 Ax. ACB are equal t to the three angles CBA, BAC, * 13. 1. ACB; but the angles ACD, ACB are equal * to

two right angles ; therefore also the angles CBA, +1 Ax. BAC, ACB, are equal t to two right angles.

Wherefore, if a side of a triangle, &c. Q. E. D.

Cor. 1.-All the interior an-
gles of any rectilineal figure, fi
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
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; *2 Cor. that is,* together with four right angles. There

fore all the 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 rectili. neal figure are together equal to four right angles.

15. 1.