PROPOSITION XXVI.

THEOREM.
If two Triangles have two Angles of the one equal

to two Angles of the otber, each to each, and
one side of the one equal to one side of the
other, either the Side lying between the equal
Angles, or which subtends one of the equal An-
gles; the remaining Sides of the one Triangle
shall be also equal to the remaining Sides of ibe
otber, each to bis correspondent Side ; and the
remaining Angle of the one equal to the remain.

ing Angle of the other. LET there be two Triangles A B C, DEF, hav,

ing two Angles A B C, B C A, of the one, equal to two Angles D EF, EFD, of the other, each to each, that is, the Angle A B C equal to the Angle DEF, and the Angle BCA equal to the Angle EF D. And let one side of the one be equal to one Side of the other, which first let 'be the Side lying between the equal Angles, viz. the Side B C equal to the Side E F. I say, the remaining Sides of the one Triangle will be equal to the remaining Sides of the other, each to each, that is, the Side A B equal to the Side D E, and the Side A C equal to the Side DF, and the remaining Angle B A C equal to the remaining Angle EDF.

For if the Side A B be not equal to the Side DE, one of them will be the greater, which let be AB, make G B equal to D Е, and join G C.

Then, because BG is equal to DE, and B C to EF, the two Sides G B, B C, are equal to the two Sides DE, EF, each to each; and the Angle G B C equal to the Angle DEF. The Base G C is * equal to the * 4 of tbisa Base D F, and the Triangle G B C to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides. Therefore the Angle GCB is equal to the Angle DFE. But the Angle D F E, by the Hypo

thefis, is equal to the Angle BCA; and so the Angle
BCG is likewise equal to the Angle B C A, the less
to the greater, which cannot be. Therefore A B is
not unequal to D Е, and consequently is equal to it.
And so the two Sides A B, BC, are equal to the two

Sides DE EF, and the Angle ABC equal to the * 4 of tbis. Angle DEF: and consequently the Base AC * is

equal to the Bafe D F, and the remaining Angle
BAC equal to the remaining Angle EDF.

Secondly, Let the Sides that are subtended by the
equal Angles be equal, as A B equal to D E. Í fay,
the remaining Sides of the one Triangle are equal to
the remaining Sides of the other, viz. A C to D F,
and B C 1 EF; and also the remaining Angle
BAC equal to the remaining Angle E DF.

For if B C be unequal to EF, one of them is the greater, which let be B C, if possible, and make BH equal to E F, and join A H.

Now, because B H is equal to E F, and A B to DE, the two Sides A B, BH, are equal to the two Sides D E, E F, each to each, and they contain equal Angles : Therefore the Base A H is equal to the Bale DF, and the Triangle A B H shall be equal to the Triangle D EF, and the remaining Angles equal to the remaining Angles, each to each, which subtend

the equal Sides; and so the Angle BHA is equal to + From the the Angle EFD. But EFD is + equal to the An

. Нур. :

gle BCA; and consequently the Angle B H A is
equal to the Angle BCA: Therefore the outward

Angle BH A of the Triangle A HC, is equal to the *.16 of this, inward and opposite Angle BCA; which is * im

possible: Whence B C is not unequal to EF; there-
fore it is equal to it. But A B is also equal to D E.
Wherefore the two Sides A B, BC, are equal to the
two Sides DE, EF, each to each ; and they contain
equal Angles. And so the Base A C is equal to the
Base DF, the Triangle BAC to the Triangle DEF,
and the remaining Angle BAC equal to the remain-
ing Angle EDF. If, therefore, two Triangles have
two Angles equal, each to each, and one side of the one
equal to one side of the other, either the Side lying between
the equal Angles, or which subtends one of the equal An-
gles; the remaining Sides of the one Triangle fall be also
equal to the remaining Sides of the other, each to its corri-

sponden

Spondent Side, and the remaining Angle of the one equal to the remaining Angle of the other ; which was to be demonstrated.

PROPOSITION XXVII.

THEOREM.

If a Right Line, falling upon two Right Lines,

makes the alternate Angles equal between them-
selves, the two Right Lines fall be parallel.
ET the Right Line E F, falling upon two Right

Lines A B, CD, make the alternate Angles A EF, E F D, equal between themselves. I say, the Right Line AB is parallel to CD.

For if it be not parallel, A B and C D, produced towards B and D, or towards A and C, will meet : Now let them be produced towards B and D, and meet in the Point G.

Then the outward Angle A EF of the Triangle GEF is * greater than the inward and opposite An- . 16 of this. gle E F G, and also equal to it; which is abfurd. + From tóc Therefore A B and CD, produced towards B and D, Hypwill not meet each other. By the same way of Reasoning, neither will they meet, being produced towards C and A. But Lines that meet each other on

Def. 35 neither Side, are & parallel between themselves. Therefore A B is parallel to C D. Therefore, if a Right Line falling upon two Right Lines, makes the alternate Angles equal between themselves, the two Right Lines fall be parallel ; which was to be demonstrated.

PRO.

PROPOSITION XXVIII.

THEOREM.

If a Right Line, falling upon two Right Lines,

makes the outward Angle with the one Line
equal to the inward and opposite Angle with
the other on the fame Side, or tbe inward An-
gles on the same Side together equal to two
Right Angles, the two Right Lines fall be
parallel between themselves.
ET the Right Line E F falling upon two Right

Lines AB, CD, make the outward Angle EGB equal to the inward and opposite Angle GHD; or the inward Angles B GH, GHD, on the fame Side together, equal to two Right Angles. I say, the Right

Line A B is parallel to the Right Line CD. . From Ibe For because the Angles E G B is * equal to the AnНур. : gle G HD, and the Angle EGB + equal to the Antis of tbis.

gle A GH, the Angle A G H Ihall be equal to the

Angle GHD; but these are alternate Angles. 1 27 of tbis. Therefore A B is f parallel to CD.

Again, because the Angles BGH, GHD, are 1 13 of this.

equal to two Right Angles, and AGH, BGH, are | equal to two Right ones, the Angles A GH, BGH, will be equal to the Angles BGH, GHD; and if the common Angle B G H be taken from both, there will remain the Angle A GH equal to the Angle GHD; but these are alternate Angles. "Therefore A B is parallel to CD. If, therefore, a Right Line, falling upon two Right Lines, makes the outward Angle with the one Line equal to the inward and opposite Angle with the other on the fame Side, or the inward Angles on the fame Side together equal to two Right Angles, the two Right Lines Mall be parallel between themselves; which was to be demonstrated.

PRO

PROPOSITION XXIX.

THEORE M.
If a Right Line falls upon two Parallels, it will

make the alternate Angles equal between them.
selves ; the outward Angle equal to the inward
and opposite Angle, on the fame Side ; and the
inward Angles on the same Side togeiber equal to

two Right Angles. L

ET the Right Line E F fall upon the parallel

Right Lines A B, CD. I say, the Alternate Angles A GH, GHD, are equal between themselves; the outward Angle E G B is equal to the inward one GHD, on the same Side ; and the two inward ones, BGH, GHD, on the same Side, are together equal to two Right Angles.

For if AG H be unequal to GHD, one of them will be the greater. Let this be AGH; then because the Angle A GH is greater than the Angle G HD, add the common Angle BGH to both : And so the Angles AGH, BGH, together, are greater than the Angles BGH, GHD, together. But the Angles AGH, BGH, are equal to two Right ones *. .13 of this. Therefore BGH, GHD, are less than two Right Angles. And so the Lines A B, C D, infinitely produced + will meet each other ; but because they are + Ax. 12. parallel they will not meet. Therefore the Angle AG H is not unequal to the Angle GHD. Wherefore it is necessarily equal to it.

But the Angle A GH is equal to the Angle I 15 of tbis. EGB: Therefore EGB is also equal to GHD

Now add the common Angle B GH; and then EGB, BGH, together, are equal to BGH, GHD, together; but EGB, and BGH, are equal to two Right Angles. Therefore also BGH, and GHD, Thall be equal to two Right Angles. Wherefore, if Right Line falls upon two Parallels, it will make the alternate Angles equal between themselves; the outward Angle equal to the inward and opposite Angle, on the farne Side; and the inward Angles on the fame Side together equai 19 two Right Angles; which was to be demonstrated.

PRO.