PROPOSITION XXVI.

THEOREM.

t

If two Triangles have two Angles of the one equal to two Angles of the other, 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, fubtends one of the equal Angles; the remaining Sides of the one Triangle fhall be alfo equal to the remaining Sides of the other, each to bis correfpondent Side; and the remaining Angle of the one equal to the remaining Angle of the other.

LET there be two Triangles A B C, DEF, hav

ing two Angles ABC, BCA, of the one, equal to two Angles DEF, EFD, of the other, each to each, that is, the Angle A B C equal to the Angle DEF, and the Angle B C A 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. 1 fay, 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 BAC equal to the remaining Angle E D F.

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

Then, because BG is equal to DE, and BC 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 DE F. The Bafe G C is equal to the * 4 of this. Bafe D F, and the Triangle GBC to the Triangle DEF, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides. Therefore the Angle G C B is equal to the Angle DFE. But the Angle DFE, by the Hypo

thefis, is equal to the Angle B C A ; and fo the Angle BCG is likewife equal to the Angle B C A, the lefs to the greater, which cannot be. Therefore A B is not unequal to DE, and confequently is-equal to it. And fo the two Sides A B, B C, are equal to the two Sides DE EF, and the Angle A B C equal to the *4 of this. Angle DEF and confequently the Base A C✶ is equal to the Bafe D F, and the remaining Angle BA C equal to the remaining Angle E D F.

Hyp.

Secondly, Let the Sides that are fubtended by the equal Angles be equal, as A B equal to DE. 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 BC to EF; and alfo the remaining Angle BAC equal to the remaining Angle EDF.

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

.

Now, because BH is equal to EF, and A B to DE, the two Sides A B, BH, are equal to the two Sides DE, E F, each to each, and they contain equal Angles: Therefore the Bafe A H is equal to the Bafe DF; and the Triangle A B H fhall be equal to the Triangle D E F, and the remaining Angles equal to the remaining Angles, each to each, which fubtend the equal Sides; and fo the Angle BHA is equal to From the the Angle EFD. But EFD is + equal to the Angle BCA; and confequently the Angle B HA is equal to the Angle BCA: Therefore the outward Angle BHA of the Triangle A H C, is equal to the 16 of this. inward and oppofitè Angle B C A; which is * impoffible: Whence BC is not unequal to E F; therefore it is equal to it. But A B is alfo equal to D E. Wherefore the two Sides A B, B C, are equal to the two Sides DE, E F, each to each; and they contain equal Angles. And fo the Bafe AC is equal to the Bafe DF, the Triangle BAC to the Triangle DEF, and the remaining Angle BAC equal to the remaining Angle ED F. 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 fubtends one of the equal Angles; the remaining Sides of the one Triangle shall be alfa equal to the remaining Sides of the other, each to its correfponden

fpondent 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-
felves, the two Right Lines fhall be parallel.

LET the Right Line EF, falling upon two Right
Lines A B, C D, make the alternate Angles
AEF, EFD, equal between themselves. I fay,
the Right Line AB is parallel to C D.

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 oppofite An- * 16 of this. gle E F G, and alfo equal † to it; which is abfurd. † From the Therefore A B and C D, produced towards B and D, Hyp. will not meet each other. By the fame way of Reafoning, neither will they meet, being produced towards C and A. But Lines that meet each other on 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 fhall be parallel; which was to be demonftrated.

‡ Def. 35°

PRO

From the

Hyp.
tis of this.

127 of this.

13 of this.

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 oppofite 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 fhall be parallel between themselves.

L

ET the Right Line E F falling upon two Right Lines A B, CD, make the outward Angle E G B equal to the inward and oppofite Angle GHD; or the inward Angles BG H, GHD, on the fame Side together, equal to two Right Angles. I fay, the Right Line A B is parallel to the Right Line C D.

*

For because the Angles E G B is equal to the Angle G H D, and the Angle E G B + equal to the Angle A GH, the Angle AGH fhall be equal to the Angle GHD; but these are alternate Angles. Therefore A B is ‡ parallel to C D.

Again, because the Angles BGH, GHD, are equal to two Right Angles, and AGH, BGH, are equal to two Right ones, the Angles AGH, BGH, will be equal to the Angles BGH, GHD; and if the common Angle BGH 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 oppofite 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 fhall be parallel between themselves; which was to be demonstrated.

PRO

PROPOSITION XXIX.

THEOREM.

If a Right Line falls upon two Parallels, it will make the alternate Angles equal between themfelves; the outward Angle equal to the inward and oppofite Angle, on the fame Side; and the inward Angles on the fame Side together equal to two Right Angles.

L

ET the Right Line E F fall upon the parallel Right Lines A B, CD. Ifay, the Alternate Angles AGH, GHD, are equal between themfelves; the outward Angle EGB is equal to the inward one GHD, on the fame Side; and the two inward ones, BGH, GHD, on the fame Side, are together equal to two Right Angles.

For if A G H be unequal to G HD, one of them will be the greater. Let this be AGH; then because the Angle AGH is greater than the Angle G HD, add the common Angle BG H to both: And fo the Angles AGH, BGH, together, are greater than the Angles BGH, G HD, together. But the Angles AGH, BGH, are equal to two Right ones*. 13 of this. Therefore B G H, GHD, are lefs than two Right Angles. And fo 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 AGH is not unequal to the Angle G HD. Wherefore it is neceffarily equal to it.

But the Angle AGH is equal to the Angle 15 of this. EG B: Therefore E GB is also equal to G H D.

Now add the common Angle BGH; and then EG B, BGH, together, are equal to BGH, GHD, together; but EG B, and BGH, are equal to two Right Angles. Therefore alfo BG H, and GHD, fhall 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 oppofite Angle, on the fame Side; and the inward Angles on the fame Side together equal to two Right Angles; which was to be demonftrated.

PRO.