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Book I. qual to AC, and let the ftraight lines AB, AC be produced to D and E, the angle ABC fhall be equal to the angle ACB, and the angle CBD to the angle BCE.
a 3. I.
b 4. I.
C 3. Ax.
In BD take any point F, and from AE, the greater, cut off AG equal to AF, the lefs, and join FC, GB.
Because AF is equal to AG, and AB to AC; the two fides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the bafe FC is equal to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal to the remaining angles of the other, each to each, to which the equal fides are oppofite; 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 D 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 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 fides are oppofite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG: And, fince it has been demonftrated, 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 bafe 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 bafe. Therefore the angles at the bafe, &c. Q. E. D.
COROLLARY. Hence every equilateral triangle is also equiangular.
PROP. VI. THEOR.
F two angles of a triangle be equal to one another, the fides alfo which fubtend, or are oppofite to, the equal angles fhall be equal to one another.
Let ABC be a triangle having the angle ABC equal to the Book I. angle ACB; the fide AB is alfo equal to the fide AC.
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. 1.
qual to AC, the lefs, and join DC; there
fore, because in the triangles DBC, ACB,
DB is equal to AC, and BC common to
equal to it Wherefore, if two angles, &c. Q. E. D.
PROP. VII. THEOR.
PON the fame bafe, and on the fame 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.
If it be poffible, let there be two triangles ACB, ADB, upon the fame bafe AB, and upon the fame fide of it, which have their fides CA, DA, terminated in the extremity A of the bafe, equal to one another, and likewife 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, becaufe AC is equal to AD, the angle ACD is equal to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the angle A ADC is greater alfo than BCD;
a j. f.
much more then is the angle BDC greater than the angle BCD. Again, becaufe 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 impoffible.
But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F, therefore, because AC is equal to AD in the triangle ACD, the angles ECD, FDC upon the other fide of the bafe CD are equal to one another; but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewife greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal A to DB, the angle BDC is equal to the angle BCD; but BDC has been proved to be greater than the fame BCD, which is impoffible. The cafe in which the vertex of one triangle is upon a fide of the other needs no demonftration.
Therefore upon the fame bafe, and on the fame fide of it, there cannot be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewise those which are terminated in the other extremity, Q. E. D.
PROP. VIII. THEOR.
two triangles have two fides of the one equal to two fides of the other, each to each, and have likewise their bafes equal; the angle which is contained by the two fides of the one fhall 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 fides AB, AC equal to the two fides DE, DF, each to each, viz. AB to DE, and AC to A
DF; and alfo the
bafe BC equal to
the base EF. The
For, if the tri
angle ABC be B
applied to DEF,
fo that the point B be on E, and the ftraight line BC upon EF; the point C fhall alfo coincide with the point F, because
BC is equal to EF; therefore BC coinciding with EF, BA and Book J. AC fhall coincide with ED and DF; for, if the bafe BC coincides with the bafe EF, but the fides BA, CA do not coincide with the fides ED, FD, but have a different fituation, as EG, FG; then, upon the fame bafe EF, and upon the fame fide of it, there can be two triangles that have their fides which are terminated in one extremity of the bafe equal to one another, and likewife their fides terminated in the other extremity: But this is impoffible; therefore. if the base BC coin- a 7.1. cides with the bafe EF, the fides BA, AC cannot but coincide with the fides ED, DF; wherefore likewife the angle BAC coincides with the angle EDF, and is equal to it. Therefore b 8, Ax. if two triangles, &c. Q. E. D.
O bifect a given rectilineal angle, that is, to divide
Let BAC be the given rectilineal angle, it is required to bi
b I. f.
Take any point D in AB, and from AC cut off AE c- 3 3. x. qual to AD; join DE, and upon it defcribe an equilateral triangle DEF; then join AF; the ftraight line AF bifects the angle BAC.
Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two fides DA, AF, are equal to the two fides EA, af, each to each; and the base DF is e- B qual to the bafe EF; therefore the
angle DAF is equal to the angle
€ 8. I.
EAF; wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done.
PROP. X. P R O B.
bifect a given finite ftraight line, that is, to divide it into two equal parts.
Let AB be the given ftraight line; it is required to divide it
into two equal parts.
Defcribe upon it an equilateral triangle ABC, and bifect the angle ACB by the ftraight line CD. AB is cut into two g. 1. equal parts in the point D.
a 3. .
c 8. I.
PROP. XI. PROB.
O draw a ftraight line at right angles to a given ftraight line, from a given point in the fame.
Let AB be a given ftraight 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 defcribe the equila
teral triangle DFE, and join
Because DC is equal to CE,
fides DC, CF are equal to the
two EC, CF, each to each; and the bafe DF is equal to the bafe EF; therefore the angle DCF is equal to the angle ECF; and they are adjacent angles. But, when the adjacent angles which one ftraight line makes with another ftraight line are 10. Def. 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 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 demonftrated, that two ftraight lines cannot have a common fegment.
If it be poffible, let the two ftraight 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