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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 “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 com-
mon to the two triangles AFC, A
AGB ; therefore the base FC is e-
qual * to the base GB, and the tri-
angle AFC to the triangle AGB;
and the remaining angles of the one
are equal" to the remaining angles
AGB: And because the whole AF is
equal to the whole AG, of which the D E
parts AB, AC are equal; the re-
mainder BF shall be 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 is
equal to the angle CGB, and the base BC is common to the two
triangles BFC, CGB; wherefore the triangles are equal b,
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, fince 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 ; the remaining angle ABC is therefore equal to the re-
maining 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 o-
ther side of the base. Therefore the angles at the base, &c.
Q. E. D.
CoRoll ARY. Hence every equilateral triangle is also equi-
- P R O P. VI. T H E O R.
r TPON the same base, and on the same fide of it, see N. there cannot be two triangles that have their fides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.
But if one of the vertices, as D, be within the other tritherefore, because AC is equal to AD F. F in the triangle ACD, the angles ECD, FDC upon the other fide of the base CD are equal “to one another; but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal B to DB, the angle BDC is equal" to the angle BCD; but BDC has been proved to be greater than the same BCD, which is impossible. 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 fide 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.
F two triangles have two fides of the one equal to two fides 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 fides equal to them, of the other.
Let ABC, DEF be two triangles having the two sides AB, AC equal to the two fides DE, DF, each to each, viz. AB to
O bisečt a given rectilineal angle, that is, to divide
it into two equal angles. -
Let AB be the given straight line; it is required to divide it
point D. Which was to be done, A o D B
Let AB be a given straight line, and C a point given in it i 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 describe" the equila-
teral triangle DFE, and join
FC ; the straight line FC drawn
from the given point C is at
right angies to the given
straight line AB. -
Because DC is equal to CE,
and FC common to the two
triangles DCF, ECF; the two A" -
jö, CF are equal to the A C E B
two EC, CF, each to each ; and the base DF is equal to the
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
equal to one another, each of them is called a right" angle ;
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. 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 segment AB common to both of them. From the point B
draw BE at right angles to AB; and because ABC is a straight