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a 16, 1.
PROP. XXVII. THEOR.
makes 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 lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.
For, if it be not parallel, AB and CD being produced thall meet either towards B,D, or towards A, C; let them be produced and meet towards B, D in the point Ġ; therefore GEF is a tria angle, and its exterior angle ACF is greater a than the interior and opposite angle EFG; but it is also equal to it, which is impossible; there.
G be demonstrated that they do C 'F
meet neither way, though $ 35. Def. produced ever so far, are parallelo to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.
PRO P. XXVIII. THE O R.
makes the exterior angle equal to the interior and opposite upon the fame side of the line ; or makes the in. terior angles upon the fame side together equal to two right angles; the two straight lines shall be parallel to one another.
Let the straight line EF, which
Because the angle EGB is e
angle Book I. ángle EGB equal to the angle AGH, the angle AGH is equal to the angle GHD; and they are the alternate angles; therefore
a IS.I. AB is parallel to CD. Again, because the angles BGH, GHD 5 27.1. are equal to two right angles, and that AGH, BGH are allo < By Hyp. equal d to two right angles; the angles AGH, BGH are equal
d 13. 1. to the angles BGH, GHD: Take away the common angle BGH, sherefore 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. E. D.
a straight line falls upon two parallel straight lines, it See the
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 lide, E. GHD , and the
two interior angles BGH, GHD upon the same fide are together equal to two right angles.
B For, if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater ; and
H D because the angle AGH is greater
F 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 to two right angles ; therefore the angles a 13. 1. 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 the same lide less than two right angles, do meet together if continually produced; therefore thé • 11. an. ftraight lines AB, CD, if produced far enough, Thall meet; but See the they never meet, fince they are parallel by the hypothesis ; this propotherefore the angle AGH is not unequal to the angle GHD, fition. that is, it is equal to it; but the angle AGH is equal o to the b 15. 1. angle EGB; therefore likewise EGB is equal to GHD; add to
Book I. each of these the angle BGH; therefore the angles EGB, BGH
are equal to the angles BGH, GHD; but EGB, BGH are eC 13. I.
qual to two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D.
Traight 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
H * to the angle GKD, and it was
F shewn that ihe angle AGK is e.
O draw a straight line through a given point paral.
a given itraight 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 to the E
A F straight line BC.
In BC take any point D, and jo'n
Because the straight line AD, which meets the two straight lincs BC, IF, makes the alternate angles EAD, ADC equal to one another, Ek is parallel to BC. Therefore the straight line
E AF is drawn through the given point A parallel to the given Book I. straight line BC. Which was to be done.
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.
b 29. I.
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 opposite angles CAB, ABC; and the three interior ang es of the triangle, viz. ABC, BCA, CAB are together equal to two right angles.
Through the point C draw CE parallel to the straight
a 31. I. line AB; and because AB is
А parallel to CE and AC mets
E them, the alternate angles BAC, ACE are equal. Again, because AB is parallel to CE, and BD falls upon them, the exterior angle ECD B
D 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 ang'e ACD is equal to the two interior and opposite angles CAB, ABO; 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 allo the c 13. 6. angles CBA, BAC, 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 rect lineai figure, together D with four right angies, are equil to tw ce as many right angles as
E the tigure has fides.
С rectilineal figure ABCDE can be divided into as
F many triangles as the figure has fides, by dr-wing Itraight lines from a point F within the figure
Book I. to each of its angles. And, by 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 Gdes of the figure ; and the same angles are equal to the angles of the fi
gure, together with the angles at the point F, which is the coma 2. Cor.
mon 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 sides.
COR. 2. All the exterior angles of any re&tilineal figure, are together equal to four right angles.
Because every interior angle
ABC, with its adjacent exterior b 13, I. ABD, is equal b to two right
angles; therefore all the interior,
PRO P. XXXIII. THEOR.
equal and parallel straight lines, towards the same
Join BC; and because AB is pa-
a 29. 1.
b 4. I.