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a 3. 1.
O 3. Ax.
Book I. qual 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 a to AF, the less, and join FC, GB.
Because AF is equal to AG, and AB to AC; the two sides FA, AC are equal to the two GA, AB, each to each ; and they contain the angle FAG common to the two triangles AFC, A
AGB ; therefore the base FC is ea b 4. I.
qual to the base GB, and the tri-
E parts AB, AC are equal; the remainder BF shall be equal to the remainder CG; and FC was proved to be equal to GB; therefore the two fides BF, FC are equal to the two CG, GB, each to each ; and the angle BFC is equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore the 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 CUG: 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 allo equal; the remaining angle ABC is therefore 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 fide of the base. Therefore the angles at the base, &c. Q. E. D.
COROLLARY. Hence every equilateral triangle is also equiangular.
.PROP. VI. THEOR.
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 Book I. angle ACB; the Gide AB is also equal to the side AC.
W 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 e- a 3. ti qual to AC, the less, and join DC; there
A fore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to D both, the two 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 baie DC is equal to the · base AB, and the triangle DBC is equal to the triangle b ACB, the less to the
b 4. 1. greater; which is abfurd. Therefore B AB is not unequal to AC, that is, it is equal to it Wherefore, if two angles, &c. Q. E. D. Cor. Hence every equiangular triangle is also equilateral.
PON the same base, and on the same side of it, See Ni
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.
If it be possible, let there be two triangles ACB, ADB, upon the same base AB, and upon the same fide of it, which have their lides CA, DA, terminated in the extremity A of the bafe, equal to one another, and like
C D wise their fides CB, DB that are terminated in B.
Join CD; then, in the cafe in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal to the angle ADC:
á š. But the angle ACD is greater than the angle BCD; therefore the angle
B 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 equal" to the angle BCD ; but it has been demonitrated to be greater than it; which is impoflible.
Book I. But if one of the vertices, as D, be within the other tri-
therefore, because AC is equal to AD
FDC upon the other side of the base $ $.8. CD are equal to one another ; but the
angle ECD is greater than the angle
greater than BCD; much more
Therefore upon the same base, and on the same fide of it,
PROP. VIII. THEOR.
sides of the other, each to each, and have likewise
Let ABC, DEF be two triangles having the two fides AB,
For, if the tri-
BC is equal to EF; therefore BC coinciding with EF, BA and Book 1. AC shall coincide with ED and DF; for, if the base BC coincides with the base EF, but the Edes BA, CA do not coincide with the fides ED, FD, but have a different situation, as EG, FG; then, upon the same base EF, and upon the same fide of it, there can be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise their fides terminated in the other extremity : But this is impoffible; therefore if the base BC coin. a 7. s. cides with the base EF, the sides BA, AC cannot but coincide with the fides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal to it. Therefore b 8. Ax. if two triangles, &c. Q. E. D.
70 bifect a given rectilineal angle, that is, to divide
it into two equal angles. Let BAC be the given rectilineal angle, it is required to bi, fect it.
Take any point D in AB, and from AC cut off AE C-13. 1. qual to AD; join DE, and upon iç describe an equilateral triangle DEF;
bl. ! then join AF; the straight line AF
' bisects the angle BAC.
Because AĎ is equal to AE, and AF is common to the two triangles D E DAF, EAF; the two sides DA, AF, are equal to the two kides EA, AF, each to each; and the base DF is e- B qual to the base EF; therefore the F angle DAF is equal to the angle EXF; wherefore the given rectilineal angle BAC is bifected by the straight line AF. Which was to be done.
C 8. I.
ho 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 * s. 1. the angle ACB by the straight line CD. AB is cut into two og.is equal parts in the point D.
a 3: B b 1. I.
Odraw a straight line at right angles to a given
straight line, from a given point in the same, See N.
Let AB be a given straight line, and C a point given in it ; it is required to draw a straight line from the point C at right angles to AB.
Take any point D in AC, and make . CE equal to CD, and upon DE describeb the equila
Because DC is equal to CE,
two EC, CF, each to each ; and the base Dr is equal to the ¢ 8. 1. base EF; therefore the angle DCF is equal to the angle ECF;
and they are adjacent angles. But, when the adjacent angles
which one straight line makes with another straight line are d 10. Def. equal to one another, each of thein is called a right d angle ;
therefore each of the angles DCF, ICF is a right angle. Wherefore from the given point C, in the given ftraight 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.
If it be possible, let the two straight lines ABC, ABD have the fegment AB common to both of them. From the point B draw BE at right angles to AB; and because ABC is a straight