And, if the angle ABC, which is common, be taken away, the remaining angle ABE will be equal to the remaining angle ABD; the lefs to the greater, which is abfurd.

The line BE, therefore, is not in the fame right line with CB; and the fame may be proved of any other line but BD; confequently CBD is one continued right line, as was to be fhewn.

PRO P. XV. THEOREM.

If two right lines interfect each other, the oppofite angles will be equal.

Let the two right lines AB, CD interfect each other in the point E; then will the angle AEC be equal to the angle DEB, and the angle AED to the angle CEB.

For, fince the right line CE falls upon the right line AB, the angles AEC, CEB, taken together, are equal to two right angles (Prop. 13.)

And, because the right line BE falls upon the right line CD, the angles BED, CEB, taken together, are alfo equal to two right angles (Prop. 13.)

The angles AEC, CEB, taken together, are, therefore, equal to the angles BED, CEB taken together (Ax. 1.)

C 3

And,

And, if the angle CEB, which is common, be taken away, the remaining angle AEC will be equal to the remaining angle BED (Ax. 3.)

And, in the same manner, it may be fhewn that the angle AED is equal to the angle CEB, Q. E. D.

COROLL. All the angles made by any number of right lines, meeting in a point, are together equal to four right angles.

PROP. XVI. THEOREM.

If one fide of a triangle be produced, the outward angle will be greater than either of the inward oppofite angles.

Let ABC be a triangle, having the fide AB produced to D; then will the outward angle CBD be greater than either of the inward oppofite angles BAC or ACB.

For, bifect BC in E (Prop. 10.), and join AE; in which, produced, take EF equal to AE (Prop. 3.), and join BF.

Then, fince AE is equal to EF, EC to EB (by Conft.) and the angle AEC to the angle BEF (Prop. 15.), the angle ACE will, alfo, be equal to the angle EBF (Prop. 4.)

But the angle CBD is greater than the angle EBF; com fequently it is also greater than the angle ACE.

And, if CB be produced to G, and AB be bisected, it

may be shewn, in like manner, that the angle ABG, or its equal CBD, is greater than CAB. Q.E. D.

The greater fide of every triangle is oppofite to the greater angle; and the greater angle to the greater fide.

Let ABC be a triangle, having the fide AB greater than the fide AC; then will the angle ACB be greater than the angle ABC.

For, fince AB is greater than AC, let AD be taken equal to AC (Prop. 3.), and join CD.

Then, fince CDB is a triangle, the outward angle ADC is greater than the inward oppofite angle DBC (Prop. 16.)

But the angle ACD is equal to the angle ADC, because AC is equal to AD; confequently the angle ACD is, also, greater than DBC or ABC.

And, fince ACD is only a part of ACB, the whole angle ACB must be much greater than the angle ABC.

Again, let the angle ACB be greater than the angle ABC, then will the fide AB be greater than the fide Ac.

For, if AB be not greater than AC, it must be either equal or lefs.

But it cannot be equal, for then the angle ACB would be equal to the angle ABC (Prop. 5.), which it is not.

Neither can it be lefs, for then the angle ACB would be less than the angle ABC (Prop. 17.), which it is not.

The fide AB, therefore, is neither equal to AC, nor less than it; confequently it must be greater, Q. E. D,

PRO P. XVIII, THEOREM,

Any two fides of a triangle, taken toge ther, are greater than the third fide.

Let ABC be a triangle; then will any two fides of it, taken together, be greater than the third fide.

For, in AC produced, take. CD equal to CB (Prop. 3.), and join BD.

Then, because CD is equal to CB (by conft.), the angle CDB will be equal to the angle CBD (Prop. 5.)

But the angle ABD is greater than the angle CBD, consequently it must also be greater than the angle ADB. And, fince the greater fide of every triangle, is op-. pofite to the greater angle (Prop. 17.), the fide AD is greater than the fide AB.

But AD is equal to AC and CB taken together (by conft.); therefore AC, CB are alfo greater than AB.

And, in the fame manner, it may be fhewn, that any other two fides, taken together, are greater than the third fide. Q. E. D,

PROP. XIX. PROBLEM.

To defcribe a triangle, whofe fides fhall be equal to three given right lines, provided any two of them, taken together, be greater than the third,

B
C

Let A, B, C be the three given right lines, any two of which, taken together, are greater than the third; it is required to make a triangle whofe fides fhall be equal to A, B, C refpectively.

Draw any right line DG; on which take DE equal to A, EF equal to B, and FG equal to c (Prop. 3.)

From the point E, with the distance ED, describe the circle KHD, cutting DG in K; and from the point F, with the diftance FG, defcribe the circle GHL, cutting DG in L.

Then, becaufe EG is greater than ED (by Hyp.), or its equal EK, the point G, which is in the circumference of the circle GHL, will fall without the circle KHD.

And, because FD is greater than FG (by Hyp.), or its equal FL, the point D, which is in the circumference of the circle KHD, will fall without the circle GHL,

But