angle ABC will co-incide with the whole Triangle DEF, and will be equal thereto; and the remaining Ax. 8. Angles will co-incide with the remaining Angles †, and will be equal to them, viz. the Angle ABC equal to the Angle DEF, and the Angle A CB equal to the Angle DFE. Which was to be demonstrated.

PROPOSITION V.

THEOREM.

The Angles at the Bafe of an Ifofceles Triangle are equal between themselves: And if the equal Sides be produced, the Angles under the Bafe fhall be equal between themfelves.

ET ABC be an Ifofceles Triangle, having the Side A B equal to the Side AC; and let the equal Sides A B, A C, be produced directly forwards to D and E. I fay the Angle A B C is equal to the Angle A CB, and the Angle CBD equal to the Angle BCE.

For affume any Point F in the Line BD, and from 3 of this AE cut off the Line AG equal to AF, and join FC, GB.

Then because A F is equal to A G, and A B to AC, the two Right Lines F A, AC, are equal to the two Lines GA, AB, each to each, and contain the com+ 4 of this. mon Angle FAG; therefore the Base F C is equal t to the Bafe G B, and the Triangle AFC equal to the Triangle AGB, and the remaining Angles of the one equal to the remaining Angles of the other, each to each, fubtending the equal Sides, viz. the Angle ACF equal to the Angle ABG; and the Angle AFC equal to the Angle AGB. And because the whole AF is equal to the whole AG, and the Part AB equal to the Part A C, the Remainder BF is equal to the Remainder CG. But FC has been proved to be equal to GB; therefore the two Sides BF, FC, are equal to the two Sides CG, GB, each to each, and the Angle B F C equal to the Angle CGB; but they have a common Bafe BC. There¬ fore alfo the Triangle BFC will be equal to the Triangle CG B, and the remaining Angles of the one

Ax. 3.

2.

equal

equal to the remaining Angles of the other, each to each, which subtend the equal Sides. And fo the Angle FBC is equal to the Angle GCB: and the Angle BCF equal to the Angle CB G. Therefore because the whole Angle ABG has been proved equal to the whole Angle ACF, and the Part CBG equal to BCF, the remaining Angle A B C will be equal to * Ax. 3. the remaining Angle A CB; but these are the Angles at the Base of the Triangle ABC. It hath likewise been proved, that the Angles F B C, G C B, under the Bafe, are equal; therefore the Angles at the Bafe of Ifofceles Triangles are equal between themselves; and if the equal Right Lines be produced, the Angles under the Base will be alfo equal between themselves.

Coroll. Hence every Equilateral Triangle is alfo Equiangular.

PROPOSITION VI.

THEOREM.

If two Angles of a Triangle be equal, then the Sides fubtending the equal Angles will be equal between themselves.

ET ABC be a Triangle, having the Angle A B C equal to the Angle ACB. I fay the Side AB is likewife equal to the Side A C.

For if AB be not equal to AC, let one of them, as AB, be the greater, from which cut off BD equal to AC+, and join DC. Then because D B is equal to † 3 of this. A C, and BC is common, DB, BC, will be equal to AC, CB, each to each, and the Angle DBC equal to the Angle ACB, from the Hypothefis ; therefore the Base DC is equal to the Base A B, and ‡ 4 of this. the Triangle DBC equal to the Triangle A CB, a Part to the Whole, which is abfurd; therefore A B is not unequal to A C, and confequently is equal to it. Therefore if two Angles of a Triangle be equal between themselves, the Sides fubtending the equal Angles are likewise equal between themselves. Which was to be demonftrated.

Coroll. Hence every Equiangular Triangle is alfo Equilateral.

PROPOSITION VII.

THEOREM.

On the fame Right Line cannot be constituted two
Right Lines equal to two other Right Lines,
each to each, at different Points, on the fame
Side, and having the fame Ends which the first
Right Lines have.

'OR, if it be poffible, let two Right Lines AD,

D B, equal to two others A C, CB, each to each, be constituted at different Points C and D, towards the fame Parts CD, and having the fame Ends A and B which the firft Right Lines have, fo that CA be equal to AD, having the fame End A which CA hath; and CB equal to DB, having the fame End B; and let CD be joined.

Then because AC is equal to AD, the Angle #5 of this. A CD will be equal to the Angle ADC, and consequently the Angle ADC is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle BCD. Again, because CB is equal to D B, the Angle BDC will be equal to the Angle BCD; but it has been proved to be much greater, which is impoffible. Therefore on the fame Right Line cannot be conftituted two Right Lines equal to two other Right Lines, each to each, at different Points, on the fame Side, and having the fame Ends which the firft Right Lines have; which was to be demonstrated.

PRO

PROPOSITION VIII.

THEOREM.

If two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafes equal, then the Angles contained under the equal Sides will be equal.

ET the two Triangles be A B C, DEF, having two Sides A B, AC, equal to two Sides DE, DF, each to each, viz. A B equal to DE, and AC to DF; and let the Bafe BC be equal to the Base EF. I fay, the Angle B A C is equal to the Angle EDF.

For if the Triangle ABC be applied to the Triangle DEF, fo that the Point B may co-incide with E, and the Right Line BC with EF, then the Point C will co-incide with F, because BC is equal to EF. And fo fince BC co-incides with EF, BA and A C will likewife co-incide with ED and DF. For if the Bafe BC fhould co-incide with E F, and at the fame Time the Sides BA, AC, fhould not co-incide with the Sides ED, DF, but change their Position, as EG, GF, then there would be conftituted on the fame Right Line two Right Lines, equal to two other Right Lines, each to each, at several Points, on the fame Side, having the fame Ends. But this is proved to be otherwise *; therefore it is impoffible for the * Sides BA, AC, not to co-incide with the Sides ED, DF, if the Bafe BC co-incides with the Bafe EF; wherefore they will co-incide, and confequently the Angle BAC will co-incide with the Angle EDF, and will be equal to it. Therefore if two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bafes equal, then the Angles contained under the equal Sides will be equal; which was to be demonftrated.

7 of this..

PRO

*

PROPOSITION. IX.

PROBLEM.

To cut a given Right-lin'd Angle into two equal

Parts.

ET BAC be a given Right-lin❜d Angle, which is required to be cut into two equal Parts.

Affume any Point D in the Right Line A B, and

3 of this cut off AE from the Line A C equal to AD* ; join + 1 of this. DE, and thereon make † the Equilateral Triangle DEF, and join A F. I fay, the Angle B A C is cut into two equal Parts by the Line A F.

For because AD is equal to A E, and AF is common, the two Sides DA, A F, are each equal to the two Sides AE, AF, and the Bafe DF is equal to 8 of this, the Base EF; therefore the Angle DAF is equal to the Angle E AF. Wherefore a given Right-lin❜d Angle is cut into two equalParts; which was to be done.

**

I of this.

PROPOSITION X.

PROBLEM.

To cut a given finite Right Line into two equal

LET a

Parts.

ET AB be a given finite Right Line, required to be cut into two equal Parts.

*

Upon it make an Equilateral Triangle A B C, and 9 of this. bifect the Angle ACB by the Right Line CD. I fay, the Right Line A B is bifected in the Point D.

For because A C is equal to CB, and CD is common, the Right Lines AC, CD, are each equal to the two Right Lines B C, CD, and the Angle ACD 4 of this. equal to the Angle BCD; therefore the Base AD, is equal to the Bale DB. And fo the Right Line AB is bifected in the Point D; which was to be done,

PRO