D

A

sides DB, BC are equal to the two AC, CB, each to
each; and the angle DBC is equal to the angle ACB;
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 it.
Wherefore, if two angles, &c. Q. E. D. B
COR.-Hence every equiangular triangle is also
equilateral.

PROP. VII. THEOR.

с

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 other extremity, equal to one another.

C D

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

Join CD; then, in the case in which the vertex of each of the tri- A

B

4. 1

angles is without the other triangle, because AC is equal to AD, the angle ACD is equal to the angle * 5. 1. ADC; but the angle ACD is greater than the angle BCD; therefore the angle ADC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, by hypothesis, the angle BDC is equal* * 5. 1. to the angle BCD; but this is impossible, because it has been proved that the angle BDC is greater than the angle BCD.

But if one of the vertices, as D, be within the other

с

* 5.1.

5. 1,

E F

triangle ACB; produce AC, AD to E, F; therefore,
because in the triangle ACD, AC is
equal to AD, the angles ECD, FDC
upon the other side of the base CD
are equal to one another; but the

*

angle ECD is greater than the angle
BCD; therefore the angle FDC is
likewise greater than BCD; much A

D

B

more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal to the angle BCD; but this is impossible, because the angle BDC has been proved to be greater than the same angle BCD. 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, 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 other extremity. Q. E. D.

PROP. VIII. THEOR.

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 A

EF; the angle BAC shall

be equal to the angle
EDF.

For, if the triangle ABC
be applied to DEF, so

DO

B

CE

F

that the point B be on E, and the straight line BC upon EF; the point C shall also coincide with the point F, because BC is equal to EF. Therefore BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if the base BC coincide with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation as EG, FG; then, upon the same base EF, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity: but this is impossible; therefore, if the base BC coincide with the base *7. 1. 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. * 8 Ax. Therefore if two triangles, &c. Q. E. D.

PROP. IX. PROB.

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.

A

Take any point D in AB, and from AC cut off* AE * 3. 1. equal to AD; join DE, and upon it describe,* on the * 1. 1. side remote from A, an equilateral triangle DEF; then join AF: the straight line AF shall bisect the angle BAC.

D

F

C

Because AD is equal to AE, and AF is common to the two triangles B 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; therefore the angle DAF is

* 8. 1.

* 1. 1.

* 9. 1.

* 4. 1.

* 3. 1.

* 1. 1.

equal to the angle EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done.

PROP. X. PROB.

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 it into two equal parts.

*

Describe upon it an equilateral triangle ABC, and
bisect the angle ACB by the straight line CD; then
AB shall be cut into two equal parts in the point D.
Because AC is equal to CB, and

CD common to the two triangles.
ACD, BCD; the two sides AC, CD
are equal to BC, CD, each to each;
and the angle ACD is equal to the
angle BCD; therefore the base AD A

D

B

Which

is equal to the base DB, and the straight line AB is
divided into two equal parts in the point D.
was to be done.

PROP. XI. PROB.

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

Let AB be a given straight line, and C a point given in it; it is required to draw from the point C a straight line at right angles to AB.

Take any point D in AC, and make CE equal to
CD, and upon DE describe the equilateral triangle
DFE, and join FC; the straight line FC drawn from
the given point C shall be
at right angles to the given
straight line AB.

Because DC is equal to
CE, and FC common to the
two triangles DCF, ECF; A

F

the two sides DC, CF are equal to the two EC, CF, each to each; and the base DF is equal to the base

录

EF; therefore the angle DCF is equal to the angle * 8. 1.

ECF; and they are adjacent angles. But, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right* angle; therefore each of the *10 Def. angles DCF, ECF, is a right angle. Therefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done.

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

E

If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight line, the angle CBE is equal to the angle EBA; in the same manner, because ABD is a straight line, the angle DBE is equal to the angle EBA; wherefore the angle DBE is equal to the angle CBE, the less to the greater; which is impossible: therefore two straight lines cannot have a common segment.

A

PROP. XII. PROB.

B

D

To draw a straight line perpendicular to a given straight line of an 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. It is required to draw from the point C a straight line perpendicular to AB.

c 2

A

E

H

F

G

B

D

*10 Def.