angle ABC will co-incide with the whole Triangle DEF, and will be equal thereto; and the remaining +4x. 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 ACB equal to the Angle DFE; which was to be demonftrated.

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 themselves.

LET ABC be an 1fofceles Triangle, having the Side A B equal to the Side AC; and let the equal Sides A B, AC, be produced directly forwards to D and E. I fay, the Angle ABC is equal to the Angle AC B, and the Angle C B D equal to the Angle BCE.

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

Then, because A F is equal to A G, and A B to A C, the two Right Lines F A, A C, are equal to the two Lines G A, A B, each to each, and contain the com† 4 of this mon Angle FAG; therefore the Base F C is equal t to the Bafe GB, and the Triangle A F C 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 A BG; and the Angle AFC, equal to the Angle A G B. And because the Whole A F is equal to the Whole A G, and the Part A B 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. Therefore alfo the Triangle BFC will be equal to the 4 of this. Triangle CGB*, and the remaining Angles of the one

I Ax. 3.

equal

equal to the remaining Angles of the other, each to each, which fubtend the equal Sides. And fo the Angle FBC is equal to the Angle GCB; and the Angle BCF equal to the Angle CBG. 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 thefe are the Angles at the Bate of the Triangle A CB. It hath likewise been proved, that the Angles FBC, GCB, under the Bafe, are equal; therefore, the Angles at the Bafe of 1fafceles Triangles are equal between themselves; and if the equal Right Lines be produced, the Angles under the Bafe will be alfo equal between themselves; which was to be demonftrated.

Coroll. Hence every Equilateral Triangle is alfo Equiangular.

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

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

For if AB be not equal to A C, let one of them, as A B, be the greater, from which cut off BD equal to AC+, and join DC. Then, becaufe BD 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 Bafe DC is equal to the Bafe A B, and 1 4 of this. the Triangle DBC equal to the Triangle A C B, 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 likewife equal between themselves; which was to be demonftrated.

Coroll.

Coroll. Hence every Equiangular Triangle is alfo
Equilateral.

PROPOSITION VII.

THEOREM.

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 first
Right Lines have.

FOR, if it be poffible, let two Right Lines AD,

D B, equal to two others A C, CB, each to each, be conftituted at different Points C and D, towards the fame Part C D, 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 D B, having the fame End B.

Cafe 1. The Point D cannot fall in the Line A C; for inftance at F: For then (AD that is) A F would not be equal to A C.

Cafe 2. If it be faid that D falls within the Triangle ABC; draw CD, and produce BD, BC, to F, and E. Now, fince A D is affirmed to be equal to A C, the Angle ADC is equal to the Angle ACD*; and confequently the Angle ACD is greater than FDC: Moreover ECD is greater than ACD, therefore ECD is much greater than FDC. But it is alfo faid, that BD is equal to BC, and so the Angle ECD under 5 of this. the Base of the Ifofceles Triangle is equal to FDC*; whereas it hath been proved to be much greater, which is abfurd: Therefore D doth not fall within the Triangle.

Cafe 3. Suppofe D fell without the Triangle ABC; join C Ď.

Then, because AC is equal to AD, the Angle +5ofibis. ACD will be equal + to the Angle A D C, and confequently the Angle ADC is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle B C D.. Again, becaufe CB is equal to D B, the Angle B DC 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 conflituted 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 demonftrated.

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.

LET the two Triangles be ABC, DEF, having two Sides, A B, AC, equal to two Sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; and let the Bafe B C 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 E F, 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 Base B C fhould co-incide with EE, and at the fame Time the Sides BA, AC, fhould not co-incide with the Sides ED, DF, but change their Pofition, 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 feveral Points, on the fame Side, having the fame Ends. But this is proved to be otherwife +; therefore it is impoffible for thet 7 of this. Sides B A, A C, not to co-incide with the Sides ED, DF, if the Base B C co-incides with the Bafe EF; wherefore they will co-incide, and confequently the Angle BAC will co-incide with the Angle EDFt, 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,

PRO

PROPOSITION IX.

PROBLEM.

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

LE

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 AC equal to A D† ; join 1 of this, DE, and thereon make the equilateral Triangle DEF, and join AF. I fay, the Angle B A C is cut into two equal Parts by the Line A F.

For, because A D is equal to A E, and AF is common, the two Sides DA, AF, are equal to the two Sides A E, AF, and the Bafe D F is equal to the 18 of this. Bafe EF; therefore the Angle DAF is equal to the Angle EAF. Wherefore, a given Right-lin'd Angle is cut into two equal Parts; which was to be done.

1 of this.

PROPOSITION X.

PROBLEM.

To cut a given finite Right Line into two equal

Parts.

LETAB be a given finite Right Line, required

to be cut into two equal Parts.

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

For, because AC is equal to CB, and C D is common, the Right Lines AC, CD, are equal to the two Right Lines B C, CD, and the Angle ACD 14 of this. equal to the Angle B CD; therefore the Bafe A D is equal to the Bafe D B. And fo, the Right Line AB is bifected in the Point D; which was to be done.

PRO