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OF EUCLID.

. For, if it be not greater, it must either be equal to it, or less; Book I. but the angle BAC

is not equal to the angle EDF, because then the base BC would be equal a to EF; but

a 4. 1. it is not; therefore the angle BAC is not equal to the angle EDF; neither is it less; because then the base BC would be less than the base

b 24. 1. EF; but it is not; therefore the angle B

СЕ F BAC is not less than the angle EDF; and it was shown 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.

PROP. XXVI. THEOR. If 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 A

D EF; the other sides shall be equal, each

G
to each, viz. AB to
DE, and AC to DF;
and the third angle
BAC to the third
angle EDF.
For, if AB be not

B
CE

F equal to DE, one of them must be the greater. Let AB be the greater of the two, and make BG equal to DE, and join GC; therefore, because BG is equal to DE, and BC to EF, the two

Book I. sides GB, BC are equal to the two DE, EF, each to each ;

and the angle GBC is equal to the angle DEF; therefore the a 4. 1.

base GC is equal a to the base DF, and 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 ; but the angle ABC is equal to the angle DEF; therefore the base 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 equal angles in each

A

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

F 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 angļes; therefore the base AH is equal to the base DF, and the triangle ABH to the triangle DEF, and the other angles are equal, each to each, to which the equal sides 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 oppob 16. 1. site angle BCA; which is impossible; wherefore 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.

PROP. XXVII. THEOR.

Book I. If a straight line falling upon two other straight lines makes the alternate angles equal to one another, these two straight lines are parallel.

Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.

For, if it be not parallel, AB and CD being produced shall meet either towards B, D, or towards A, C; let them be produced and meet towards B, D in the point G; therefore GEF is a triangle, and its exterior angle AEF is greater a than the a 16. 1. interior and opposite angle EFG; but it is also equal to it, which is impossible; A

B
therefore AB and CD being
produced do not meet to-
wards B, D. In like man-

may
be demonstra- C

F

D ted, that they do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel to one another. AB therefore is parallel b 35. Def. to CD. Wherefore, if a straight line,&c. Q. E. D.

ner it

PROP. XXVIII. THEOR.

If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line ; or makes the interior angles upon the same side together equal to two right angles ; the two straight lines are parallel to one another.

Let the straight line EF, which E falls upon the two straight lines AB, CD, make the exterior

G angle EGB equal to the interior A

-B and opposite angle GHD upon the same side; or make the interior angles on the same side C

-D BGH, GHD together equal to

H two right angles; AB is parallel to CD.

F Because the angle EGB is equal to the angle GHD, and

a 15. 1. b 27. 1.

Book I. also a to the angle AGH, the angle AGH is equal to the

angle GHD; and they are the alternate angles; therefore AB

is parallel b to CD. Again, because the angles BGH, GHD c By Hyp.

are equal o to two right angles; and AGH, BGH are also d 13. 1. equal d to two right angles; the angles AGH, BGH are equal

to the angles BGH, GHD: Take away the common angle BGH, therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles; therefore AB is parallel to CD. Wherefore, " if a straight line," &c. Q. E. D.

PROP. XXIX. THEOR.

See the notes on

this propo

If a straight line fall upon two parallel straight

lines, it makes the alternate angles equal to one ansition. other ; and the exterior angle equal to the interior

and opposite upon the same side; and likewise the two interior angles upon the same side together equal to two right angles.

Let the straight line EF fall upon the parallel straight lines AB, CD; the alternate angles AGH, GHD are equal to one another; and the exterior angle EGB is equal to the interior and opposite, upon the same side, E GHD; and the two interior angles BGH, GHD upon the same side are together equal to two A G

B right angles.

For, if AGH be not equal to
GHD, one of them must be great-

С
H

Ꭰ .
er than the other; let AGH be the
greater; and because the angle

F AGH is greater than the angle GHD, add to each of them the angle BGH; therefore the angles AGH, BGH are greater than the angles BGH, GHD; but the angles AGH, BGH are equal a to two right angles; therefore the angles BGH, GHD are less than two right angles; but those straight lines which, with another straight line falling upon them, make

the interior angles on the same side less than two right angles, * 12. ax. do meet * together if continually produced; therefore the

straight lines AB, CD, if produced far enough, shall meet; but they never meet, since they are parallel by the hypothesis ;

therefore the angle AGH is not unequal to the angle GHD, b 15. 1. that is, it is equal to it; but the angle AGH is equal to the

angle EGB; therefore likewise EGB is equal to GHD; add

a 13. 1.

See the notes on

this propo

sition.

to each of these the angle BGH; therefore the angles EGB, Book I. BGH are equal to the angles BGH, GHD; but EĞB, BGH. are equal o to two right angles; therefore also BGH, GHD - 13. 1. are equal to two right angles. Wherefore, " if a straight,&c. Q. E. D.

PROP. XXX. THEOR.

a 29. 1.

Straight lines which are parallel to the same straight line are parallel to one another.

Let AB, CD be each of them parallel to EF, AB is also parallel to CD.

Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AB, EF, the angle AGH is equal a to the angle GHF. Again, because the straight line A

B GK cuts the parallel straight lines EF, CD, the angle GHF

H H

F is equal a to the angle GKD; and it was shown that the angle C

K

D AGK is equal to the angle GHF; therefore also AGK is equal to GKD; and they are alternate angles ; therefore AB is parallel to CD. Where- b 27. 1. fore, “ straight lines," &c. Q. E. D.

PROP. XXXI. PROB.

To draw a straight line through a given point parallel to a given straight line.

Let A be the given point, and BC the given straight line ; it is required to draw a straight line through the point A, parallel E

A F to the straight line BC.

In BC take any point D, and join AD; and at the point A, in the straight line AD, make a the

angle

. B D С DAĚ equal to the angle ADC; and produce the straight line EA to F.

Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel to BC. I'herefore the straight line b 27. 1.

a 23. 1.

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