PROPOSITION XV.

THEOREM.

If two Right Lines mutually cut each other, the oppofite
Angles are equal.

L

ET the two Right Lines AB CD mutually cut each other in the Point E. Ifay, the Angle AEC is equal to the Angle DEB; and the Angle CEB equal to the Angle AED.

*

For becaufe the Right Line A E, ftanding on the Right Line CD, makes the Angles CEA, AED: *13 of this. These both together fhall be equal to two Right Angles. Again, because the Right Line DE ftanding upon the Right Line AB, makes the Angles AED, DEB: Thefe Angles together are* equal to two Right Angles. But it has been prov'd, that the Angles CEA, AED, are likewife together equal to two Right Angles. Therefore the Angles CEA, AED, are equal to the Angles AED, DEB. Take away the Common 4x.3. Angle AED, and the Angle remaining CEA, is † equal to the Angle remaining BED. For the fame Reafon, the Angle CEB fhall be equal to the Angle DEA. Therefore if two Right Lines mutually cut each other, the oppofite Angles are equal; which was to be demonstrated.

Coroll. 1. From hence it is manifeft, that two Right
Lines mutually cutting each other, make Angles
at the Section equal to four Right Angles.
Coroll. 2. All the Angles conftituted about the fame
Point, are equal to four Right Angles.

PRO

PROPOSITION XVI.

THEORE M.

If one Side of any Triangle be produced, the outward
Angle is greater than either of the inward Oppofite
Angles.

LE

ET ABC be a Triangle, and one of its Sides BC, be produced to D. I fay, the outward Angle ACD is greater than either of the inward Angles CBA, or BAC.

Forbifect AC in E*, and join BE, which produce 10 of this, to F, and make EF equal to BE. Moreover, join

FC, and produce AC to G.

Then, becaufe AE is equal to EC, and BE to EF, the two Sides AE, EB, are equal to the two Sides CE, EF, each to each, and the Angle A EB

† equal to the Angle FEC; for they are oppofite +15 of this, Angles. Therefore the Bafe AB, is equal to the 4 of this. Bale FC; and the Triangle A EB, equal to the Triangle FEC; and the remaining Angles of the one, equal to the remaining Angles of the other, each to each, fubtending the equal Sides. Wherefore the the Angle BAE, is equal to the Angle ECF; but the Angle ECD, is greater than the Angle ECF; therefore the Angle ACD, is greater than the Angle BAE. After the fame manner, if the Right Line BC, be bifected, we demonftrate that the Angle BCG, that is, the Angle ACD, is greater than the Angle ABC. Therefore one Side of any Triangle being produced, the outward Angle is greater than either of the inward oppofite Angles; which was to be demonftrated.

PROPOSITION XVII.

THEOREM.

Two Angles of any Triangle together, howsoever taken, are less than two Right Angles.

LET ABC be a Triangle. I fay, two Angles of it together, how foever taken, are less than two Right Angles.

C

For

For produce BB to D.

*

Then because the outward Angle A CD of the *16 of this. Triangle ABC, is greater than the inward oppofite Angle ABC: If the common Angle ACB be added, the Angles ACD, ACB, together, will be greater than the Angles ABC, ACB together: But ACD, +13 of this. ACB, are fequal to two Right Angles. Therefore ABC, BCA, are less than two Right Angles. In the fame manner we demonftrate that the Angles. BAC, ACB, as alfo CAB, ABC, are lefs than two Right Angles. Therefore two Angles of any Triangle together, howsoever taken, are less than two Right Angles; which was to be demonstrated.

PROPOSITION. XVIII.

THEOREM.

The greater Side of every Triangle fubtends the greater

LE

Angle.

ET ABC be a Triangle, having the Side AC greater than the Side AB. I fay the Angle ABC is greater than the Angle BCA.

For because AC is greater than AB, AD may be made equal to AB, and BD be join'd,

Then because ADB is an outward Angle of the 16 of this. Triangle BDC, it will be greater than the inward. 5 of this, oppofite Angle DCB. But ADB is equal to ABD; because the Side AB is equal to the Side AD. Therefore the Angle ABD is likewife greater than the Angle ACB; and confequently ABC fhall be much greater than ACB. Wherefore the greater Side of every Triangle fubtends the greater Angle, which was to be demonftrated.

PRO

For produce BB to D.

*

Then because the outward Angle A CD of the *16 of this. Triangle ABC, is greater than the inward oppofite Angle ABC: If the common Angle ACB be added,

the Angles ACD, ACB, together, will be greater than the Angles ABC, ACB together: But ACD, +13 of this. ACB, are equal to two Right Angles. Therefore ABC, BCA, are less than two Right Angles. In the fame manner we demonftrate that the Angles BAC, ACB, as alfo CAB, ABC, are lefs than two Right Angles. Therefore two Angles of any Triangle together, howsoever taken, are less than two Right Angles; which was to be demonstrated.

PROPOSITION. XVIII.

THEOREM.

The greater Side of every Triangle fubtends the greater
Angle.

LET ABC be a Triangle, having the Side AC
greater than the Side AB. I fay the Angle ABC
is greater than the Angle BCA.

For because AC is greater than AB, AD may be made equal to AB, and BD be join'd,

Then because ADB is an outward Angle of the 16 of this. Triangle BDC, it will be greater than the inward, ts of this, oppofite Angle DCB. But ADB is equal to ABD; because the Side AB is equal to the Side AD. Therefore the Angle ABD is likewife greater than the Angle CB; and confequently ABC fhall be much greater than ACB. Wherefore the greater Side of every Triangle fubtends the greater Angle, which was to be demonftrated.

PRO