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 CËB. 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 opposite 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 fhewn, in like manner, that the angle ABG, or its equal CBD, is greater than CAB. Q. E. D.

triangle is oppo

The greater fide of every triangle is fite 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, alfo, 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.

be equal to the angle GAB, and the angle ABC to the angle ABG (Prop. 4.)

But the triangles AGB, DFE, are identical; confequently the angles of the triangle DFE will, also, be equal to the corresponding angles of the triangle acb. Q. E.D.

Let ABC, DEF be each of them right angles; then will ABC be equal to DEF.

For conceive the angle DEF to be applied to the angle ABC, fo that the point E may coincide with the point B, and the line ED with the line BA.

And if EF does not coincide with BC, let it fall, if poffible, without the angle ABC, in the direction BG; and produce AB to H.

Then, because the angles ABC, ABG are right angles (by Hyp.), the lines CB, GB will be each perpendicular to AH (Def. 8. 9.)

And, fince a right line which is perpendicular to another right line, makes the angles on each fide of it equal (Def. 8.), the angle CBA will be equal to the angle CBH, and the angle GBA to the angle GBH.

But the angle GBA is greater than the angle CBA, or its equal CBH; confequently the angle GBH is alfo greater

than the angle CBH; that is, a part is greater than the whole, which is abfurd.

The line EF, therefore, does not fall without the angle ABC and in the fame manner it may be fhewn that it does not fall within it; confequently EF and BC will coincide, and the angle DEF be equal to the angle ABC, as was to be fhewn.

PRO P. IX. PROBLEM.

To bisect a given rectilineal angle, that is, to divide it into two equal parts.

Let BAC be the given rectilineal angle; it is required to divide it into two equal parts.

Take any point D in AB, and from AC cut off ae equal to AD (Prop. 3.), and join DE.

Upon DE describe the equilateral triangle DFE (Prop. 1.), and join AF; then will AF bifect the angle BAC, as was required.

For AD is equal to AE, by conftruction; DF is alfo equal to FE (Def. 16.), and AF is common to each of the triangles AFD, AFE.

But when the three fides of one triangle are equal to the three fides of another, each to each, the angles which are oppofite to the equal fides are, also, equal (Prop. 7.)

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