For produce B C to D.

Then because the outward Angle ACD 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. A CB, are † equal to two Right Angles. Therefore ABC, BCA, are less than two Right Angles.

* 16 oft bis. +5 of this

In

the fame manner we demonftrate that the Angles BAC, ACB, as alfo CAB, ABC, are less than two Right Angles. Therefore two Angles of any Triangle together, bowfoever taken, are less than two Right Angles; which was to be demonftrated.

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 B CA.

For because AC is greater than AB, AD may be made equal to AB, and BD be joined.

*

Then because ADB is an outward Angle of the Triangle BDC, it will be greater than the inward oppofite Angle DCB. But ADB ist equal to ABD; because the Side A B is equal to the Side AD. Therefore the Angle ABD is likewife greater than the Angle ACB; and confequently A B C fhall be much greater than A CB. Wherefore the greater Side of every Triangle fubtends the greater Angle; which was to be demonstrated.

PRO

PROPOSITION XIX.

THEOREM.

The greater Angle of every Triangle fubtends the greater Side.

LE

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

For if it be not greater, AC is either equal to AB, or less than it. It is not equal to it, because then the

Angle ABC would be equal to the Angle ACB;* 5 of this, but it is not. Therefore AC is not equal to AB; neither will it be lefs; for then the Angle ABC would bet less than the Angle ACB; but it is not. There- † 18 of this. fore AC is not lefs than AB. But likewise it has been proved not to be equal to it: Wherefore AC is greater than AB. Therefore the greater Angle of every Triangle fubtends the greater Side; which was to be demonftrated.

PROPOSITION XX.

THEOREM.

Two Sides of any Triangle, bowfoever taken, are together greater than the third Side.

LET ABC be a Triangle: I fay two Sides thereof, howfoever taken, are together greater than the third Side, viz. the Sides BA, AC, are greater than the Side BC; and the Sides AB, BC, greater than the Side AC, and the Sides BC, CA, greater than the Side AB.

For produce BA to the Point D, so that AD be *equal to AC, and join DC.

*

3 of thisa

Then because DA is equal to AC, the Angle ADC fhall be equal to the Angle ACD. But the Anglet 5 of this, BCD is greater than the Angle ACD. Wherefore the Angle BCD is greater than the Angle ADC; and becaufe DCB is a Triangle, having the Angle BCD greater than the Angle BDC, and the greater Angle

* 19 of this. Angle fubtends * the greater Side; the Side DB will be greater than the Side BC. But DB is equal to BA and AC together. Wherefore the Sides BA, AC, together, are greater than the Side BC. In the fame Manner we demonftrate, that the Sides AB, BC, together, are greater than the Side CA; and the Sides BC, CA, together, are greater than the Side AB. Therefore two Sides of any Triangle, howsoever taken, are together greater than the third Side; which was to be demonstrated.

If two Right Lines be drawn from the extreme Points of one Side of a Triangle to any Point within the fame, these two Lines fhall be less than the other two Sides of the Triangle, but contain a greater Angle.

FOR

OR let two Right Lines BD, DC, be drawn from the Extremes B, C, of the Side B C of the Triangle ABC, to the Point D within the fame. I fay BD, DC, are less than BA, AC, the other two Sides of the Triangle, but contain an Angle BDC greater than the Angle BAC.

For produce BD to E.

*

Then because two Sides of every Triangle together 20 of this are greater than the third, BA, AE, the two Sides of the Triangle ABE, are greater than the Side BE. Now add EC, which is common, and the Sides † Ax. 4. BA, AC, will be † greater than BE, EC.

Again, because CE, ED, the two Sides of the Triangle CED, are greater than the Side CD, add DB, which is common, and the Sides CE, EB, will be greater than CD, DB. But it has been proved, that BA, AC, are greater than BE, EC: Wherefore BA, AC, are much greater than BD, DC. Again, because 16 of this. the outward Angle of every Triangle, is greater than the inward and oppofite one: BDC, the outward Angle of the Triangle CDE, fhall be greater than the Angle CED. For the fame Reafon CEB the outward Angle of the Triangle ABE, is likewise greater