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+ Constr. 1 Ax.

centre G, at the distance GH, describe * another circle * 3 Post.
HLK; and join KF, KG: the triangle KFG shall
have its sides equal to the three straight lines A, B, C.

Because the point Fis the centre of the circle DKL,
FD is equal* to FK; but FD is equalt to the straight * 15 Def.
line A; therefore FK is equal to A: again, because
G is the centre of the circle LKH, GH is equal* to * 15 Def.
GK; bút GH is equal to C; therefore also GK is
equal to C: and FĠ is equal to B: therefore the + Constr.
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.

PROP. XXIII. PROB.

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* 22. 1.

At a given point in a given straight line, to make a recti

lineal 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
line AB, that shall be equal to the
given rectilineal angle DCE.
In CD, CE, take any points D,

14

EFA
E, and join DE; and make* the
triangle AFG, the sides of which
shall be equal to the three straight lines CD, DE, EC,
so that CD be equal to AF, ČE to AG, and DE to
FG: the angle FAG shall be equal to the angle DCE.

Because DC, CE are equal to FA, AG, each to
each, and the base DE to the base FG; the angle
DCE is equal * to the angle FAG. Therefore at the * 8. 1.
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.

B

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If two triangles have two sides of the one equal to two sides See N.

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

* 23. 1.

* 3. 1.

D

Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two DE, DF, each to each, viz. AB equal to DE, and AC to DF; but the angle BAC greater than the angle EDF: the base BC shall be 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 equal**to AC or DF,

and join EG, GF. + Hyp. Because AB is equalt to DE, and ACť to DG, the + Constr.

two sides, BA, AC are equal to the two ED, DG, each

to each, and the angle BAC is + Constr. equalt to the angle EDG; there* 4.1.

fore the base BC is equal* to the
base EG. And because DG is
equal to DF, the angle DFG is
equal* to the angle DGF; but
the angle DGF is greatert than the angle EGF;
therefore the angle DFG is greater than EGF; there-
fore much more is the angle EFG greater than the
angle EGF: and because the angle EFG of the trian-
gle EFG is greater than its angle EGF, and that the
greater* angle is subtended by the greater side; there-
fore the side EG is greater than the side EF: but EG
was proved to be equal to BC; therefore BC is greater
than EF. Therefore, if two triangles, &c. Q. E. D.

G

B

* 5. 1.

F

+ 9 Ax.

* 19. 1.

PROP. XXV. THEOR.

See N.

If 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 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 BC greater than the base EF: the angle BAC shall be greater than the angle EDF.

For, if it be not greater, it must either be equal to it, or less than it: but the angle BAC is not equal to the angle EDF, because then the base BC would be equal*

* 4. 1.

D

* 24. 10

to EF: but it ist not: therefore

+ Hyp. 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 EF; but it is + not; therefore the angle BAC

F + Hyp. 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.

B

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А.

D

G

B

E

If two triangles have two angles of the 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,
each to each, 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 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 greater than the other.

Let AB be the greater of the two, and make BG equalt to DE, and join GC: + 3. 1. therefore, because BG is equal to DE, and BC+ to + Hyp. EF, the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equalt to + Hyp. the angle DEF; therefore the base GC is equal* to the * 4. 1. 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 equalt to the angle BCA, the + 1 Ax. less to the greater, which is impossible: therefore AB

+ Hyp.

+ Hyp. * 4. 1.

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+ 3. 1. + Hyp.

is not unequal to DE, that is, it is equal to it: and BC is equalt to EF; therefore the two AB, BC are equal to the two DE, EF, each to each; and the angle ABC is equalt 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
triangle be equal to one another,
viz. AB to DE: likewise in this
case, the other sides shall be

Іi СЕ
equal, AC to DF, and BC to EF;
and also the third angle BAC

to the third angle 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 equalt angles; therefore the base AH is equalt to the base DF, and the triangle ABH 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 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 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 containt 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.

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Q. E, D.

PROP. XXVII. THEOR.

If a straight line falling upon two other straight lines

make the alternate angles equal to one another, these two straight lines shall be 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 shall be parallel to CD.

For, if it be not parallel, AB and CD being pro

B В

duced will 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* than the interior and opposite * 16. 1.
angle EFG; but it is also equalt to

+ Hyp:
it, which is impossible; therefore,

А. AB and CD being produced do not meet towards B, D. In like manner it may be demonstrated, 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 an- * 35 Def. other : therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D.

G

PROP. XXVIII. THEOR.

B

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If a straight line falling upon two other straight lines

make the exterior angle equal to the interior and oppo-
site upon the same side of the line ; or make the inte-
rior angles upon the same side together equal to two
right angles; the two straight lines shall be parallel to
one another.

Let the straight line EF, which falls upon the two
straight lines, AB, CD, make the ex-
terior angle EGB equal to the interior
and opposite angle GHD upon the
same side; or make the interior angles
on the same side BGH, GHD toge-
ther equal to two right angles: AB
shall be parallel to CD.

Because the angle EGB is equalt to the angle GHD, + Hyp.
and the angle EGB is equal* to the angle AGH, there- * 15.1.
fore the angle AGH is equalt to the angle GHD: and +1 Ax.
they are the alternate angles; therefore AB is parallel* * 27. 1.
to CD.*Again, because the angles BGH, GHD are
equal* to two right angles, and that AGH, BGH are * By Hyp.
also equal* to two right angles; therefore the angles * 13. 1.
AGH, BGH are equal+ to the angles BGH, GHD: +1 Ax.
take away the common angle BGĦ; therefore the re-
maining angle AGH is equalt to the remaining angle + 3 Ax.
GHD. and they are alternate angles; therefore AB is
parallélt to CD. Wherefore, if a straight line, &c. † 27. 1.
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