PRO P. V. THEOREM.

The angles at the base of an ifofceles triangle are equal to each other.

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Let ABC be an isofceles triangle, having the fide Ca equal to the fide CB; then will the angle CAB be equal to the angle CBA.

For, in CA and Cв produced, take any two equal parts CD, CE (Prop. 3.), and join the points AE, BD:

Then, because the two fides CA, CE of the triangle CAE, are equal to the two fides CB, CD of the triangle CBD, and the angle c is common, the fide AE will also be equal to the fide BD, the angle CAE to the angle CBD, and the angle D to the angle E (Prop. 4.)

And fince the whole CD is equal to the whole CE (by Conft.), and the part CA to the part Cв (by Hyp.), the remaining part AD will also be equal to the remaining part BE (Ax. 3.)

Hence, the two fides DA, DB, of the triangle DAB, being equal to the two fides EB, EA of the triangle EBA, and the angle D to the angle E, the angle ABD will alfo be equal to the angle BAE (Prop. 4.)

And if from the equal angles CAE, CBD, there be taken the equal angles BAE, ABD, the remaining angle

CAB

CAB will be equal to the remaining angle CBA (Ax. 3.) Q.E. D.

COROLLARY. Every equilateral triangle is also equi. angular.

If two angles of a triangle be equal to each other, the fides which are oppofite to them will also be equal.

Let ABC be a triangle, having the angle CAB equal to the angle CBA; then will the fide CA be equal to the fide CB.

For if CA be not equal to CB, one of them must be greater than the other; let CA be the greater, and make AD equal to BC (Prop. 3.) and join BD.

Then, because the two fides AD, AB, of the triangle ADB, are equal to the two fides BC, BA, of the triangle ACB, and the angle DAB is equal to the angle CRA (by Hyp.), the triangle ADB will be equal to the triangle ACB (Prop. 4.), the lefs to the greater, which is abfurd.

The fide CA, therefore, cannot be greater than the fide CB; and, in the fame manner, it may be fhewn that it cannot be lefs; confequently they are equal to each other. Q.E.D.

COROL. Every equiangular triangle is also equilateral.

PROP. VII. THEOREM.

If the three fides of one triangle be equal to the three fides of another, each to each, the angles which are oppofite to the equal fides will also be equal.

Let ABC, DEF be two triangles, having the fide AB equal to the fide DE, AC to DF, and BC to EF; then will the angle ACB be equal to the angle DFE, BAC to EDF, and ABC to DEF.

For, let the triangle DFE be applied to the triangle ACB, so that their longeft fides, DE, AB, may coincide, and the point F fall at G; and join CG.

Then, fince the fide AC is equal to the fide DF, or AG (by Hyp.), the angle ACG will be equal to the angle AGC (Prop. 5.)

And, because the fide BC is equal to the fide EF, or BG (by Hyp.), the angle BCG will be equal to the angle BGC (Prop. 5.)

But fince the angles ACG, BCG are equal to the angles AGC, BGC, each to each, the whole angle ACB will be equal to the whole angle AGB (Ax. 8.)

And, because ac is equal to AG, BC to BG, and the angle ACB to the angle AGB, the angle CAB, will, also,

be

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, alfo, 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 side 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 bifect 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 defcribe the equilateral triangle DFE (Prop. 1.), and join AF; then will AF bisect the angle BAC, as was required.

For AD is equal to AE, by construction; DF is also 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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