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are greater than BE, to each of these add EC, therefore the sides Book I. BA, AC are greater than BE, А. EC. again, because the two fides CE, ED of the triangle

D CED are greater than CD, add DB to each of these; therefore the sides CE, EB are greater than CD, DB. but it has been

B Mewn that BA, AC, are greater ihan BE, EC; much more then are BA, AC greater than BD, DC.

Again because the exterior angle of a triangle is greater than the interior and opposite angle, the exterior angle BDC of the triangle CDE is greater than CED. for the same reason, the exterior, angle CEB of the triangle ABE is greater than BAC. and it has been demonstrated that the angle BDC is greater than the angle CEB; much more then is the angle BDC greater than the angle BAC. therefore if from the ends of, &c. Q. E. D.

PROP. XXII. PROB

See N.

a. 20.1.

2. 3. 1.

O make a triangle of which the sides fhall be equal

to three given straight lines; but any two whatever of these must be greater than the thirda.

Let A, B, C be the three given straight lines, of which any two whatever are greater than the third, viz. A and B greater than C; A and C greater than B; and B and C than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each.

Take a straight line DE terminated at the point D, but unlimited towards E, and make a DF equal to A, FG to B, and GH equal to C; and from the center F, at the distance FD describe b the

b. 3. Poft. D circle DKL, and from the center G, at the distance

L GH defcribeb another

A circle HLK, and join KF, KG. the triangle KFG has its sides equal to the three straight lines A, B, C. Because the point F is the center of the circle DKL, FD B 3

qual

e

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Book I. qual o to FK; but FD is equal to the straight line A; therefore MFK is equal to A. again, because G is the center of the circle LKH, s, is. Def. GH is equal to GK; but GH is equal to C, therefore also GK

is equal to C. and FG is equal to B; therefore the three straight lines KF, FG, GK are equal to the three A, B, C. and therefore the triangle KFG has its three sides KF, FG, GK equal to the three given straight lines A, B, C. Which was to be done.

A

PROP. XXIII. PROB.
T a given point in a given straight line to make a

rectilineal angle equal to a given rectilineal angle.
Let AB be the given straight line, and A the given point in it.
and DCE the given rectilineal angle; it is required to make an
angle at the given point
A in the given straight

С

А
line AB that mail be
equal to the given reco
tilineal angle DCE.

Take in CD, CE, a-
ny points D, E, and join

D

E
DE; and make a the

F

G
triangle AFG the sides
of which shall be equal

B
to the three fuaight lines CD, DE, EC, so that CD be equal to AF,
CE to AG, and DE to FG. and because DC, CE are equal to
FA, AG, each to each, and the base DE to the base FG; the
angle DCE is equal o to the angle FAG. therefore at the given
point A in the given straight line AB, the angle FAG is made e-
qual to the given rectilineal angie DCE. Which was to be done.

PROP. XXIV. THEOR.
F two triangles have two sides of the one equal to two

sides of the other, each to each, but the angle con-
tained by the two sides of one of them greater than the
angle contained by the two kides equal to them, of the o.
ther; the base of that which has the greater angle shall
be greater than the base of the other.
Let ABC, DEF be two triangles which have the two fides AB,

AC

6. 8. 8.

IF

See N.

AC equal to the two DE, DF, each to each, viz. AB equal to DE, Book I. and AC to DF; but the angle BAC greater than the angle EDF. the base BC is also greater than the base EF.

Of the two sides DE, DF let DE be the side which is not greater than the other, and at the point D in the straight line DE make* a. 23. 1. the angle EDG equal to the angle BAC; and make DG equal bb. 3. 1. to AC or DF, and join EG, GF.

Because AB is equal to DE, and AC to DG, the two fides
BA, AC are equal to the two ED, DG, each to each, and the angle
BAC is equal to the
angle EDG; therefore A

D the base BC is equal

to the base EG, and becaufe DG is equal to DF, the angle DFG is equal to the angle DGF; but the angle

E DGF is greater than B the angle EGF, there

F fore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. and because the angle EFG of the triangle EFG is greater than its angle EGF, and that the greater e fide is opposite to the greater angle; the e. 19. 1, side EG is therefore greater than the fide EF. but EG is equal to BC; and therefore also BC is greater than EF. therefore if two triangles, &c. Q. E. D.

c. 4.

d. s.r.

PROP. XXV. THEOR.

Ftwo triangles have two sides of the one equal to two

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greater than the base of the other; the angle also contained by the sides of that which has the greater base, shall be greater than the angle contained by the sides equal to them, of the other,

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the base CB greater than the base EF. the angle BAC is likewise greater than the angle EDF.

For

B 4

2.4.1.

Book I. For if it be not greater, it must either be equal to it, or less. mbut the angle BAC is not equal to the angle EDF, because then

the base BC would be
equal to EF. but it A

D
is not; therefore the
anglc BAC is not e-
qual to

the angle
EDF. neither is it
less; because then the

base BC would be less b.24. 1. b than the base EF; B

с E but it is not; therefore the angle BAC is not less than the angle EDF. and it was fhewn that it is not equal to it; therefore the angle BAC is greater than the angle IDF. Wherefore if two triangles, &c. Q. E. D.

PROP. XXVI. THEOR.
IF
F two triangles have two angles of one equal to two

angles of the other; each to each, and one side equal to one side, viz. cither the sides adjacent to the equal angles; or the sides oppolite to equal angles in each; then shall the other sides be equal, each to each, and also the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, viz. ABC to DEF, and BCA to EFD; also one side equal to one side ; and first, let those sides be equal which are adjacent to the angles that are equal in the two triangles, viz. BC to EF. the other sides shall be equal, each to each, viz. AB to DE,

and

A
AC to DF; and the

D
third angle BAC to G
the third angle EDF.

For if AB be not
equal to DE, one of
them must be the
greater. Let AB be

с E the greater of the two, and make BG equal to DE, and join GC. therefore because BG is

equal

equal to DE, and BC to EF, the two sides GB, BC are equal to the Book I. two DE, EF, each to each; and the angle GBC is equal to the angle DEF; therefore the base GC is equal • to the base DF, and the tri-a. 4. 1. angle GBC to the triangle DEF, and the other angles to the other angles, each to each, to which the equal sides are opposite; therefore the angle GCB is equal to the angle DFE; but DFE is, by the hypothesis, equal to the angle BCA; wherefore also the angle BCG is equal to the angle BCA, the less to the greater, which is impossible. therefore AB is not unequal to DE, that is, it is equal to it. and BC is equal to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equal to the angle DEF, the base therefore AC is equal to the base DF, and the third angle BAC to the third angle EDF.

Next, let the sides which are opposite to A

D equal argles in each triangle be equal to one another, viz. AB to DE; likewise in this case, the other sides Mall be equal, AC to DF, and BC to B HC E

F EF; and also the third angle BAC to the third EDF.

For if BC be not equal to EF, let BC be the greater of them, and make BH equal to EF, and join AH. and because BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each; and they contain equal angles; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles shall be equal, each to each, to which the equal fides are opposite. therefore the angle BHA is equal to the angle EFD. but EFD is equal to the angle BCA ; therefore also the angle BHA is equal to the angle BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and opposite angle BCA; which is impossibleb, wherefore BC is not unequal to EF, that is, b. 16. 1. it is equal to it; and AB is equal to DE; therefore the two AB, BC are equal to the two DE, EF, each to each; and they contain equal angles; wherefore the base AC is equal to the base DF, and the third angle BAC to the third angle EDF. therefore if two triangles, &c.

AN

Q. E. D.

PROP.

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