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PROPOSITION 2. PROBLEM.

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; [Post. 1. and on it describe the equilateral triangle DAB,

K

[I. 1.

H

A

B

lines

and produce the straight
DA, DB to E and F. [Post. 2.
From the centre B, at the dis-
tance BC, describe the circle
CGH, meeting DFat G. [Post. 3.
From the centre D, at the dis-
tance DG, describe the circle
GKL, meeting DE at L. [Post. 3.
AL shall be equal to BC.

Because the point B is the centre of the circle CGH,

BC is equal to BG.

[Definition 15.

And because the point D is the centre of the circle GKL,

ᎠᏞ is equal to ᎠᏩ ;

and DA, DB parts of them are equal;

[Definition 15.

[Definition 24.

therefore the remainder AL is equal to the remainder BG.

[Axiom 3.

But it has been shewn that BC is equal to BG; therefore AL and BC are each of them equal to BG. But things which are equal to the same thing are equal to one another. [Axiom 1.

Therefore 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. Q.E.F. PROPOSITION 3. PROBLEM.

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, of which

AB is the greater: it is required to cut off from AB, the greater, a part equal to C the less.

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Because the point A is the centre of the circle DEF,

AE is equal to AD.

But C is equal to AD.

[Definition 15. [Construction.

Therefore AE and Care each of them equal to AD.
Therefore AE is equal to C.

[Axiom 1.

Wherefore from AB the greater of two given straight lines a part AE has been cut off equal to C the less. Q.E.F.

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If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely 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, namely, AB to DE, and AC to

DF, and the angle 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 shall be equal, each to each, to which the equal

B

sides are opposite, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

For if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB on the straight line DE, the point B will coincide with the point E, because AB is equal to DE.

[Hyp.

And, AB coinciding with DE, AC will fall on DF, because the angle BAĆ is equal to the angle EDF.

B

[Hypothesis.

Therefore also the point C will coincide with the point F, because AC is equal to DF.

[Hypothesis. But the point B was shewn to coincide with the point E, therefore the base BC will coincide with the base EF; because, B coinciding with E and C with F, if the base BC does not coincide with the base EF, two straight lines will enclose a space; which is impossible. [Axiom 10. Therefore the base BC coincides with the base EF, and is equal to it.

[Axiom 8. Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it.

[Axiom 8.

And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE.

Wherefore, if two triangles &c. Q.E.D.

PROPOSITION 5. THEOREM.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another.

Let ABC be an isosceles triangle, having the side AB equal to the side 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 BĈE.

In BD take any point F,

and from AE the greater cut off AG equal to AF the less, [1.3.

and join FC, GB.

Because A Fis equal to AG, [Constr. and AB to AC, [Hypothesis.

the two sides FA, ACare equal to the
two sides GA, AB, each to each; and
they contain the angle FAG common
to 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 one to the remaining
angles of the other, each to each, to D
which the equal sides are opposite,

B

C

namely the angle ACF to the angle ABG, and the angle AFC to the angle AGB.

[I. 4.

And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal,

[Hypothesis. the remainder BF is equal to the remainder CG. [Axiom 3. And FC was shewn 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 BFC was shewn to be equal to the angle CGB; therefore the triangles BFC, CGB are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle FBC to the angle GCB, and the angle BCF to the angle CBG.

[I. 4.

And since it has been shewn that the whole angle ABG is equal to the whole angle ACF,

and that the parts of these, 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. [Axiom 3.

And it has also been shewn that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base.

Wherefore, the angles &c. Q.E.D.

Corollary. Hence every equilateral triangle is also equiangular.

PROPOSITION 6. THEOREM.

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 AC shall be equal to the side AB.

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For if AC be not equal to AB, of them must be greater than the other.

Let AB be the greater, and from it cut off DB equal to AC the less,

and join DC.

Then, because in the triangles DBC, ACB,
DB is equal to AC,

and BC is common to both,

B

[I. 3.

[Construction.

the two sides DB, BC are equal to the two sides AC, CB, each to each;

and the angle DBC is equal to the angle ACB ; [Hypothesis. 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.

[I. 4.

[Axiom 9.

Therefore AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles &c. Q.E.D.

Corollary. Hence every equiangular triangle is also equilateral.

PROPOSITION 7. THEOREM.

On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another.

If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB, having their sides CA, DA,

which are terminated at the extremity A of the base, equal

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