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are greater than BE. to each of these add EC, therefore the sides Rook I. BA, AC are greater than BE,

А.
El. again, because the two sides
CE, ED of the triangle CED

D
are greater than CD, add DB to
each of these; therefore the sides
CZ, EB are greater than CD,
D. but it has been shewn that
BA, AC are greater than BE, B
EC ; much more then are BA, AC greater than BD), DC.

Agan 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 deinonstrated 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.

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PRO P. XXII. PRO B. To make a triangle of which the sides fliall be equal see N.

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

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; Annd 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 unli. mited towards E, and make * DF equal to A, FG to B, and GH equal to C; and from the center F at the distance FD describe to the D circle DKL. and from the

G center G, at the distance GH describe b another circle HLK, and join KF,KG. the triangle KFG has its fides equal to the three straight lines A, B, C.

Because the point F is the center of the circle DKL, FD is

a. 3. !.

b.

3.

Pont,

HE

Book I. equal 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, $ 15. Def. GH is equal to GK; 'but GH is equal to C, therefore allo 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.

E

12. I.

PROP. XXIII. PRO B.
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

А.
line AB that shall be
equal to the given recti-
lineal angle DCE.

Take in CD, CE, any
points D, E,and join DE;
and make a the triangle
AFG the sides of which

B
shall be equal to the
three straight 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 b to the angle FAG. therefore at the given point A
in the given straight line AB, the angle FAG is made equal to
the given rectilineal angle DCE. Which was to be done.

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

two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other ; the base of that which has the greater angle fhall be greater than the base of the other.

Let ABC, DEF be two triangles which have the two sides AB,

b. 8. s.

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. m 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 the angle EDG equal to the angle BAC; and make DG a. 23. 1. equal b to AC or DF, and join EG, GF.

b. 3, . Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to

CA the angle EDG; there. fore the base BC is equal to the base EG. and because DG is equal to DF, the angle DFG is equal - to the

d. s. b. angle DGF; but the

angle DGF is greater

С E than the angle EGF,

F therefore 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 side is opposite to the greater angle; the e. 19. 1. side EG is therefore greater than the side 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. S.

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F two triangles have two sides of the one equal to

two sides of the other, each to each, but the base of the one 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 fides 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.

Book 1.

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

the base BC would be
8. 4. I.

equal to EF. but it
is not ; therefore the

A
angle BAC is not e-

D
qual to the angle EDF.
neither is it less; be-
cause then the base BC

would be less b than
B. 24. 1.

the base EF ; but it is
not; therefore the an B

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

PRO P. XXVI. THEO R.

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. either the sides adjacent to the equal angles, or the sides opposite 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

D
AC to DF; and the
third angle BAC to the
third angle EDF,

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

B
CE

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

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equal to DE, and BC to EF, the two sides GB, BC are equal to Book I. the two DE, EF, each to each ; and the angle GBC is equal to the angle DEF; therefore the base GC is equal a to the base DF, anda. 4. I. the triangle 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 a to the base DF, and the third angle BAC to the third angle EDF.

Next, let the sides which are opposite to &qual angles in each A.

D triangle be equal to one another, viz. AB to DE ; likewise in this case, the other fides shall be equal, AC to DF, and BC to EF; and also the B HC L

E 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 equil sides are opposite. therefore the angle BHA is equal to the ai gle 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, where- b. 16.1. fore BC is not unequal to EF, that is, 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. Q. E. D.'

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