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PROPOSITION I. PROBLEM.
To describe an equilateral triangle upon a given finite

straight line.
Let AB be the given straight line; it is required to de-
scribe an equilateral triangle upon AB.

From the centre A, at the distance AB, describe * the circle BCD, and

3 Postu

B E late. from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines * CA, CB, to the points *1 Post. A, B; ABC shall be an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal * to AB; and because the point B is the 15 Defi

nition, centre of the circle ACE, BC is equal to BA: but it has been proved that CA is equal to AB; therefore CA, CB, are each of them equal to AB: but things which are equal to the same thing are equal* to one another;

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iom. therefore CA is equal to CB: wherefore CA, AB, BC are equal to one another: and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done.

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*

PROP. II. PROB.
From a given point to draw a straight line equal to a

given straight line.
Let A be the given point, and BC the given straight
line; it is required to draw from the point A a straight
line equal to BC.

From the point A to B draw* the straight line AB; * 1 Post. and upon it describe* the equilateral triangle DAB, 1. 1. and produce* the straight lines DA, DB, to E and F; * 2 Post. from the centre B, at the distance BC describe* the cir- * 3 Post. cle CGH, and from the centre D, at the distance DG describe the circle GKL. AL shall be equal to BC.

Because the point B is the centre of the circle CGH, BC is equal* to BG;

15 Def. and because D is the centre of the cir. cle GKL, DL is equal to DG; and +

+ Constr. DA, DB, parts of them, are equal; therefore the remainder AL is equal to the remainder* BG: but it has been shewn,

* 3 Ax.

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that BC is equal to BG; wherefore AL and BC are

each of them equal to BG: and things that are equal † 1 Ax. to the same thing are equal+ to one another; therefore

the

straight line AL is equal to BC. Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done.

PROP. III. PROB.

A

* 2.1.

From the greater of two given straight lines to cut off a

part equal to the less. Let AB and C be the two given straight lines, whereof AB is the

greater. It is required to cut off from AB,

Е В the greater, a part equal to C, the less.

From the point A draw* the straight

line AD equal to C; and from the centre A, and at the * 3 Post.

distance AD describe * the circle DEF: AE shall be equal to C.

Because A is the centre of the circle DEF, AE is + 15 Def. equal+ to AD; but the straight line C is likewise equal + Constr.

tot AD; whence AE and Care each of them equal to * 1 Ax. AD; wherefore the straight line AE is equal to* C, and

from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.

PROP. IV. THEOR.

If two triangles have two sides of the one equal to two

sides of the other, each to each ; and have likewise the angles contained by those sides equal to one another ; they shall likewise have their bases, or third sides, equal ; and the two triangles shall be equal, and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.

Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and the angle

B CE BAC equal to the angle EDF: the base BC shall be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles

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to which the equal sides are opposite, shall be equal each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE.

For, if the triangle ABC be applied to DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal t to DE: and AB coinciding with + Hyp. DE, AC shall coincide with DF, because the angle BAC is equal t to the angle EDF; wherefore also the + Hyp. point C shall coincide with the point F, because the straight line AC+ is equal to DF: but the point B was + Hyp. proved to coincide with the point E: wherefore the base BC shall coincide with the base EF; because, the point B coinciding with E, and C with F, if the base BC does not coincide with the base EF, two straight lives would inclosea space, which is impossible*. There- * 10 Ax. fore the base BC coincides with the base EF, and therefore is equalt to it. Wherefore the whole triangle ABC + 8 As. coincides with the whole triangle DEF, and is equal to it; and the other angles of the one coincide with the remaining angles of the other, and are equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, their bases shall likewise be equal, and the triangles shall be equal, and their other angles to which the equal sides are opposite shall be equal, each to each. Which was to be demonstrated.

PROP. V. THEOR.

The angles at the base of an isosceles triangle are equal

to one another; and if the equal sides be produced, the
angles upon the other side of the base shall be equal.

Let ABC be an isosceles triangle, of which the side A B is equal to AC, and let the straight lines AB, AC be produced to D and E: the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD take any point F, and from AE the greater, cut off AG equal* to AF, the less, and join FC, GB.

Because AF is equal to + AG, and AB to $ AC, the + Constr. two sides FA, AC are equal to the two GA, AB, each | Hyp. to each : and they contain the angle FAG common to

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* 3. 1,

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the two triangles AFC, AGB; therefore
the base FC is equal * to the base GB,
and the triangle AFC to the triangle
AGB; and the remaining angles of the В,
one are equal * to the remaining angles
of the other, each to each, to which the
equal sides are opposite; viz. the angle
ACF to the angle ABG, and the angle AFC to the an-
gle AGB: and because the whole AF is equal to the
whole AG, of which the parts AB, AC are equal; the
remainder BF is equal* to the remainder CG: and FC
was proved to be equal to GB; therefore the two sides
BF, FC are equal to the two CG, GB, each to each;
and the angle BFC was proved to be equal to the angle
CGB, and the base BC is common to the two triangles
BFC, CGB; wherefore these triangles are equal *, and
their remaining angles each to each, to which the equal
sides are opposite: therefore the angle FBC is equal to
the angle GCB, and the angle BCF to the angle CBG.
And, since it has been demonstrated, that the whole an-
gle ABG is equal to the whole ACF, the parts of which,
the angles CBG, BCF are also equal; therefore the
remaining angle ABC is equal + to the remaining angle
ACB, which are the angles at the base of the triangle
ABC: and it has also been proved that the angle FBC
is equal to the angle GCB, which are the angles upon
the other side of the base. Therefore the angles at the
base, &c.

COROLLARY.--Hence every equilateral triangle is also equiangular.

† 3 Ax.

Q. E. D.

PROP. VI. THEOR.

If two angles of a triangle be equal to one another, the

sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle ACB: the side AB shall be equal to the side AC.

For, if AB be not equal to AC, one of them is greater than the other: let AB be the greater; and from it cut * off DB equal to AC, the less, and join DC: therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides, DB, BC are equal to the two AC, CB, each to

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4. 1.

each; and the angle DBC is equal to the anglet ACB; + Hyp. therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle *ACB, the less to the greater, which is absurd. Therefore AB is not unequal to AC, that is, it is equal to

с it. Wherefore, if two angles, &c. Q. E. D.

COR.—Hence every equiangular triangle is also equilateral.

B

PROP. VII. THEOR.

Upon the same base, and on the same side of it, there can- See N. not be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.

If it be possible, upon the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA terminated in the extremity A of the base equal to one another, and likewise their sides, CB, DB, that are terminated in B.

Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal † to AD, the angle ACD is

B + Hyp. to the angle ADC: but the angle ACD is greater than the angle BCD; therefore the +9 Ax. angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal+ to DB, the angle BDC is + Hyp. equal to the angle BCD; but it has been demonstrated - 5. 1. to be greater than it, which is impossible.

But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F: therefore, because AC is equal † to AD in the tri

+ Hyp. angle ACD, the angles ECD, FDC EF upon the other side of the base CD are equal to one another: but the angle ECD is greater † than the angle

+ 9 Ax. BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle

equal *

• 5. 1.

*

* 5. 1.

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