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and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the base GB, (1. 4.)

and the triangle AFC is equal to the triangle AGB,

also the remaining angles of the 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 angle AGB.

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

therefore the remainder BF is equal to the remainder CG; (ax. 3.)
and FC has been proved to be equal to GB;
hence because the two sides BF, FC are equal to the two CG,
GB, each to each;

and the angle BFC has been proved to be equal to the angle CGB, also the base BC is common to the two triangles BFC, CGB;

wherefore these triangles are equal, (1. 4.)

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 angle 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;

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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.

Q. E.D.

COR. Hence an equilateral triangle is also equiangular.

PROPOSITION VI. THEOREM.

If two angles of a triangle be equal to each other, 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 A CB. Then the side AB shall be equal to the side AC.

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For, if AB be not equal to AC,

one of them is greater than the other. If possible let AB be greater than AC;

and from BA cut off BD equal to CA the less, (1. 3.) and join DC. Then, in the triangles DBC, ABC,

because DB is equal to AC, and BC is common to both triangles, 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; (hyp.) therefore the base DC is equal to the base AB, (1. 4.) and the triangle DBC is equal to the triangle ABC, the less equal to the greater, which is absurd. (ax. 9.) Therefore AB is not unequal to AC, that is, AB is equal to AC. Wherefore, if two angles, &c.

Q. E.D.

COR. Hence an equiangular triangle is also equilateral.

PROPOSITION VII. THEOREM,

Upon the same base, and on the same side of it, there cannot 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 ather extremity.

If it be possible, on 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.

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First. When the vertex of each of the triangles is without the other triangle.

Because AC is equal to AD in the triangle ACD, therefore the angle ADC is equal to the angle ACD; (1. 5.) but the angle ACD is greater than the angle BCD; (ax. 9.) therefore also the angle ADC is greater than BCD; much more therefore is the angle BDC greater than BCD. Again, because the side BC is equal to BD in the triangle BCD, (hyp.) therefore the angle BDC is equal to the angle BCD; (1. 5.) but the angle BDC was proved greater than the angle BCD, hence the angle BDC is both equal to, and greater than the angle BCD; which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle ACB.

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Produce AC to E and AD to F, and join CD.

Then because AC is equal to AD in the triangle ACD,

therefore the angles ECD, FDC upon the other side of the base.

CD, are equal to one another; (1. 5.)

but the angle ECD is greater than the angle BCD; (ax. 9.)
therefore also the angle FDC is greater than the angle` BCD;

much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD in the triangle BCD, therefore the angle BDC is equal to the angle BCD; (1. 5.) but the angle BDC has been proved greater than BCD, wherefore the angle BDC is both equal to, and greater than the angle BCD;

which is impossible.

Thirdly. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration.

Therefore, upon the same base and on the same side of it, &c. Q. E. D.

PROPOSITION VIII. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one, shall be equal to the angle contained by the two sides equal to them, of the other.

Let ABC, DEF be two triangles, having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and also the base BC equal to the base EF.

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Then the angle BAC shall be equal to the angle EDF.

For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF;

then because BC is equal to EF, (hyp.)

therefore the point C shall coincide with the point F;

wherefore BC coinciding with EF,

BA and AC shall coincide with ED, DF;

for, if the base BC coincide with the base EF, but the sides BA, AC, do not coincide with the sides ED, DF, but have a different situation as EG, GF:

Then, upon the same base, and upon the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those sides which are terminated in the other extremity;

but this is impossible. (1. 7.)

Therefore, if the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. (ax. 8.)

Therefore, if two triangles have two sides, &c. Q.E.D.

PROPOSITION IX. PROBLEM.

To bisect a given rectilineal angle, that is, to divide it into two equal angles.

Let BAC be the given rectilineal angle.
It is required to bisect it.

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In AB take any point D;

from AC cut off AE equal to AD, (1. 3.) and join DE;
on the side of DE remote from A,

describe the equilateral triangle DEF (1. 1.) and join AF.
Then the straight line AF shall bisect the angle BAC.
Because AD is equal to AE, (constr.)

and AF is common to the two triangles DAF, EAF;
the two sides DA, AF, are equal to the two sides EA, AF,
each to each;

and the base DF is equal to the base EF; (constr.)

therefore the angle DAF is equal to the angle EAF. (1. 8.) Wherefore the angle BAC is bisected by the straight line AF. Q.E.F.

PROPOSITION X. PROBLEM.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line.

It is required to divide AB into two equal parts.
Upon AB describe the equilateral triangle ABC; (1. 1.)

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and bisect the angle ACB by the straight line CD meeting AB in the point D. (1. 9.)

Then AB shall be cut into two equal parts in the point D.
Because AC is equal to CB, (constr.)

and CD is common to the two triangles ACD, BCD; the two sides AC, CD are equal to the two BC, CD, each to each; and the angle ACD is equal to BCD; (constr.)

therefore the base AD is equal to the base BD. (1. 4.) Wherefore the straight line AB is divided into two equal parts in the point D. Q. E. F.

PROPOSITION XI. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given straight line, and C a given point in it.

It is required to draw a straight line from the point C at right angles to AB.

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In AC take any point D, and make CE equal to CD; (1. 3.) upon DE describe the equilateral triangle DEF, (I. 1.), and join CF. Then CF drawn from the point C shall be at right angles to AB. Because DC is equal to EC, and FC is common to the two triangles DCF, ECF;

the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; (constr.)

therefore the angle DCF is equal to the angle ECF: (1. 8.)

and these two angles are adjacent angles.

But when the two adjacent angles which one straight line makes with another straight line, are equal to one another, each of them is called a right angle: (def. 10.)

Therefore each of the angles DCF, ECF is a right angle. Wherefore from the given point C, in the given straight line AB, FC has been drawn at right angles to AB.

Q. E. F.

COR. By help of this problem, it may be demonstrated that two straight lines cannot have a common segment.

If it be possible, let the segment AB be common to the two straight lines ABC, ABD.

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From the point B, draw BE at right angles to AB; (1. 11.)
then because ABC is a straight line,

therefore the angle ABE is equal to the angle EBC. (def. 10.) Similarly, because ABD is a straight line,

therefore the angle ABE is equal to the angle EBD;
but the angle ABE is equal to the angle EBC,

wherefore the angle EBD is equal to the angle EBC, (ax. 1.)
the less equal to the greater angle, which is impossible.
Therefore two straight lines cannot have a common segment.

PROPOSITION XII. PROBLEM.

To draw a straight line perpendicular to a given straight line of unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced any length both ways, and let C be a point without it.

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