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■ 22.1.

b 8.1.

a 23. 1.

b 3.1.

c 4. 1.

d 5. 1.

points DE; join DE, and make the
triangle AFG whose three right lines
are equal to the three right lines CD,
DE, EC. So that AF be equal to
CD, AG to CE, and FG to DE. There-
fore because the two DC, CE, are
equal to the two FA, AG, each to

Да

E

D

B

each, and the base DE equal to the base FG, the angle DCE will be equal to the angle FAG. Therefore to a given right line AB, and to a point in it, a rectilineal angle has been placed equal to the given rectilineal angle DCE. Q. E. F.

PROPOSITION XXIV.

THEOREM.*

If two triangles have two sides equal to two sides, each to each, but the angle contained by the equal sides of the one greater than the angle contained by the equal sides of the other; then shall the base of that which has the greater angle be greater than the base 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. the side AB equal to the side DE, and the side AC equal to DF. But the angle BAC greater than the angle EDF. The base BC will

be greater than the base EF. A
For because the angle BAC is
greater than the angle EDF;
to the right line DE and to the
point D in it, make the angle
EDG equal to BAC; also put B
DC equal to either AC, or

E

D

F

G

DF, and join GE, FG. Therefore because AB is equal to DE, and AC to DG; the two BA, AC, are equal to the two ED, DG, each to each; and the angle BAC is equal to the angle EDG; therefore the base BC is equal to the base EG. Again, because DG is equal to DF, the angle DFG is equal to the angle DGF; therefore the angle DFG will be greater than the angle EGF, much more will the angle EFG be greater than the angle EGF. And

* The student is not to conclude from hence that the area of that triangle which has the greater base is greater than the area of the other, for it can be clearly proved, by means of some following propositions, that its area may be either equal to the other triangle, or less than it.

e

because EFG is a triangle having the angle EFG greater than the angle EGF, also the greater side subtends the greater angle, the side EG will be greater than the side * 19. 1. EF. But the side EG is equal to the side BC. Therefore BC will also be greater than EF. If therefore two triangles have two sides, &c. Q. E. D.

PROPOSITION XXV.

THEOREM.*

If two triangles have two sides equal to two sides each to each; but the base of the one greater than the base of the other; then shall the angle contained by the equal sides of the one be greater than the angle which is contained by the equal sides 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. the side AB equal to the side DE, and the side AC to the side DF. But the base BC greater than the base EF. The angle BAC is greater than the angle EDf.

AA

E F

For if it be not greater, it is either equal or less. But the angle BAC is not equal to the angle EDF; for then the base BC would be equal to the base EF.a But it is a 4.1. not. The angle BAC is not therefore equal to the angle EDF. But neither is it less; for then the base BC would also be less than the base EF. But it is not. ↳ 24. 1. Therefore the angle BAC is not less than the angle EDF. And it was shown that it is not equal. Therefore the angle BAC is greater than EDF. If therefore two triangles, &c. Q. E. D.

PROPOSITION XXVI.
THEOREM.

If two triangles have two angles equal to two angles, each to each, and one side equal to one side, either the side which is adjacent to the equal angles, or the side which subtends one of the equal angles, they will have the remaining sides equal to the remaining sides each to each, and the remaining angle to the remaining angle.

Let ABC, DEF, be two triangles, which have the two angles ABC, BCA, equal to the two angles DEF, Efd,

* Direct demonstrations of this are given by Menelaus, Alexandrinus, and Hero.

a 4. 1.

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A

HC

E

F

each to each, viz. the angle
ABC equal to the angle DEF,
and the angle BCA equal to the
angle EFD. Let them also
have one side equal to one side,
and first that which is adjacent,
to the equal angles, viz. the
side BC to the side EF. They will also have the remain-
ing sides equal to the remaining sides, each to each, viz.
the side AB to the side DE, and the side AC to DF, and
the remaining angle BAC equal to the remaining angle
EDF. For if AB is unequal to DE, one of them is the
greater. Let AB be the greater, and make GB equal to
DE; and join GC. Therefore because BG is equal to
DE, and BC to EF; the two GB, BC, are equal to the 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 triangle GBC to the triangle DEF;
also the remaining angles equal to the remaining angles,
each to each, to which the equal sides are opposite.
Therefore the angle GCB is equal to the angle DFE; but
the angle DFE is 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. But 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. Therefore the base AC is equal
to the base DF, and the remaining angle BAC is equal
to the remaining angle EDF; but let the sides, which
subtend the equal angles, be equal to one another, as
AB to DE. Then again the remaining sides are equal
to the remaining sides; viz. AC is equal to DF, also BC
to EF; and, as before, the remaining angle BAC is
equal to the remaining angle EDF. For if BC be
unequal to EF, one of them is the greater. Let Bс be
the greater, if it be possible, and make BH equal to EF,
and join AH. Wherefore 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 DFE; also the remaining
angles will be equal to the remaining angles, each to
each, to which the equal sides are opposite. Therefore
the angle BHA is equal to the angle EFD. But EFD is

b

equal to the angle BCA.b And therefore the angle BHA By hyp. is equal to the angle BCA; the exterior angle BHA of the triangle AHC is equal to the interior and opposite angle BCA, which is impossible. Wherefore BC is not 16. 1. unequal to EF; that is, it is equal to it. But 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 triangle BAC to the triangle DEF; also the remaining angle BAC is equal to the remaining angle EDF. If therefore two triangles, &c. Q. E. D.

Deductions.

1. To draw a right line through a given point so as to make equal angles with two right lines given in position. 2. If any two triangles have the three angles of the one respectively equal to the three angles of the other; also if the perpendiculars from the vertical angles to the bases be equal, then shall the three sides of the one triangle be equal to the three sides of the other, viz. those which are opposite to the equal angles.

PROPOSITION XXVII.

THEOREM.*

If a right line falling upon two right lines makes the alternate angles equal to one another, the right lines will be parallel.

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F

E B

D

Let the right line EF falling upon the two right lines AB, CD, make the alternate angles AEF, EFD, equal to one another; the right line AB is parallel to CD. For if it is not parallel, AB, CD, being produced will meet either towards the parts B, D, or towards the C parts A, c. Let them be produced, and meet towards the parts B, D, in the point G. the exterior angle AEF of the triangle GEF than the interior and opposite angle EFG; but it is also equal; which is impossible. Therefore AB, CD, being produced, do not meet towards the parts BD. like manner we may demonstrate they do not meet

Therefore
is greater

a a

In

* The student must understand that the lines are in the same plane, otherwise the alternate angles might be equal; and the lines might or might not be parallel, as the commentary of Proclus upon this proposition fully evinces.

b

16. 1.

By hyp.

towards the parts A, C. But those right lines which being produced meet towards neither parts are parallel. Therefore AB is parallel to CD. Wherefore a right line,

&c.

Q. E. D.

Deduction.

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a

b 15. 1.

c 27.1.

PROPOSITION XXVIII.

THEOREM.

If a right line falling upon two right lines make the exterior angle equal to the interior and opposite angle towards the same parts, or the interior angles towards the same parts equal to two right angles; the right lines will be parallel to one another.

E

Let the right line EF falling upon the two right lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD, or the interior angles towards the same parts BGH, GHD, equal to two right angles. The right line AB is parallel to the right line c For because the angle EGB

CD.

d

A

G

B

D

H

F

By hyp. is equal to the angle GHD, and the angle EGB to the angle AGH,b the angle AGH will also be equal to the angle GHD; and they are alternate angles. Therefore AB is parallel to CD. Again, because the angles BGH, GHD, are equal By hyp. to two right angles, and the angles AGH, BGH, are equal to two right angles; therefore the angles AGH, BGH, will be equal to the angles BGH, GHD. away BGH, which is common. Therefore the remainder AGH is equal to the remainder GHD; and they are alternate angles. Therefore AB will be parallel to CD. therefore a right line, &c. Q. E. D.

e 13. 1.

e

Take

If

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