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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 t, and

will be equal to them, viz. the Angle A B C equal to the Angle D E F, and the Angle A CB equal to the Angle ĎFE. Which was to be demonstrated.

PROPOSITION V.

THEOREM.
The Angles at the Base of an Isosceles Triangle are

equal between themselves : And if the equal
$ides be produced, the Angles under the Base
Mall be equal between themselves.

Side A B equal to the Side AC; and let the equal Sides AB, AC, be produced directly forwards to D and E. I say the Angle A B C is equal to the Angle ACB, and the Angle CBD equal to the Angle BCE.

For affume any Point F in the Line BD, and from * 3 of tbis. A E cut off the Line A G equal * to AF, and join

FC, GB.

Then because AF is equal to AG, and A B to AC, the two Right Lines FA, AC, are equal to the two

Lines GA, A B, each to each, and contain the com+ 4 of this. mon Angle FAG; therefore the Base F C is equal †

to the Base G B, and the Triangle AF 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 An-
gle ACF equal to the Angle A BG; and the Angle
AF C equal to the Angle A GB. 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 I 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 Base BC. There
fore also the Triangle BFC will be equal to the
Triangle CGB, and the remaining Angles of the one

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equal to the remaining Angles of the other, each to
each, which subtend the equal Sides. And so the An-
gle F B C 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 ABC will be * equal to
the remaining Angle ACB; but these are the Angles
at the Base of the Triangle A BC. It hath likewise
been proved, that the Angles FBC, GCB, under the
Base, are equal; therefore the Angles at the Base of
Isosceles Triangles are equal between themselves; and
if the equal Right Lines be produced, the Angles under
the Base will be also equal between themselves.

* Ax. 3.

Coroll. Hence every Equilateral Triangle is also Equi

angular.

PROPOSITION VI.

.

THEOREM.
If two Angles of a Triangle be equal, then the sides

fubtending the equal Angles will be equal be-
tween themselves.

ET ABC be a Triangle, having the Angle

ABC equal to the Angle ACB. I say the Side
A B is likewise 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 B D equal to
ACt, and join DC. Then because D B is equal to t 3 of this.
AC, 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 Hypothesis ;
therefore the Base DC is equal to the Base A B, and I 4 of this.
the Triangle DBC equal to the Triangle ACB, a
Part to the Whole, which is abfurd; therefore A B is
not unequal to AC, and consequently 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 demonstrated.

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Coroll

. Hence every Equiangular Triangle is also Equilateral.

PROPOSITION VII.

THEOREM.
On the Same Right Line cannot be constituted two

Right Lines equal to two other Right Lines,
each to each, at different Points, on the same
Side, and having the same Ends which the first
Right Lines bave.

FOB

OR, if it be possible, let two Right Lines AD,

DB, equal to two others A C, CB, each to each, be constituted at different Points C and D, towards the same Parts CD, and having the same Ends A and B which the first Right Lines have, so that CA be equal to AD, having the fame End A which CA hath; and CB equal to D B, having the same 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 con

sequently the Angle ADC is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle B CD. 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 impossible. Therefore on the same Right Line cannot be constituted two Right Lines equal to two ather Right Lines, each to each, at different Points, on the same Side, and having the fame Ends which the first 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 Bases equal, then the Angles contained under
the equal Sides will be equal.
ET the two Triangles be A BC, DEF, having

DF, each to each, viz. A B equal to DE, and AC
to DF; and let the Base B C be equal to the Base
EF. I say, the Angle BAC is equal to the Angle
EDF.

For if the Triangle ABC be applied to the Trian-
gle D EF, so that the Point B may co-incide with E,
and the Right Line B C with EF, then the Point C
will co-incide with F, because BC is equal to EF.
And so since B C co-incides with EF, B A and AC
will likewise co-incide with ED and DF. For if
the Base B C should co-incide with EF, and at the
same Time the Sides BA, AC, should not co-incide
with the Sides E D, DF, but change their Position,
as E G, GF, then there would be constituted on the
fame Right Line two Right Lines, equal to two other
Right Lines, each to each, at several Points, on the
same Side, having the same Ends. But this is proved
to be otherwise * ; therefore it is impossible for the *
Sides BA, AC, not to co-incide with the Sides ED,
DF, if the Base B C co-incides with the Base EF;
wherefore they will co-incide, and consequently 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 Šides of the other,
each to each, and the Bases equal, then the Angles con-
tained under the equal Sides will be equal; which was
to be demonstrated.

of thisa

PRO

PROPOSITION. IX.

PROBLEM.

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

Parts.

L

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 tbis. cut off A E from the Line AC equal to AD*; join + 1 of tbis. D E, and thereon make † the Equilateral Triangle

DEF, and join AF. I say, the Angle BAC is cut into two equal Parts by the Line AF.

For because A D is equal to A E, and A F is common, the two Sides D A, AF, are each equal to the

two Sides A E, AF, and the Base D F is equal to 1 8 of this, the Base E F; therefore I 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.

PROPOSITION X.

PROBLEM.

To cut a given finite Right Line into two equal

Parts.

L
ET AB be a given finite Right Line, required to

be cut into two equal Parts. I of this.

Upon it make * an Equilateral Triangle A B C, and te of tbis. bisect † the Angle ACB by the Right Line CD. I

say, the Right Line A B is bifected in the Point D.

For because AC is equal to CB, and CD is common, the Right Lines AC, CD, are each equal to

the two Right Lines BC, CD, and the Angle ACD | 4 of tbii. equal to the Angle BCD; therefore I the Base AD,

is equal to the Bale DB. And fo the Right Line AB is bisected in the Point D; which was to be done,

PRO

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