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greater than the angle contained by the sides equal to them, of the other.

Let ABC, DEF, be two As, having the sides AB, AC = sides DE, DF, ea. to ea., and base BC > base EF; then / BAC > < EDF.

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BAC must be either >, =, or < ≤ EDF.

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and it has been proved / BAC #▲ EDF,

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EDF.

PROP. XXV. THEOR. 26. 1Eu.

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, equal to the third angle of the other.

CASE I. When sides adj. to equals in each be equal.

In As ABC, DEF,

BCEF,

let ABC / DEF,

▲ BCA = / EFD,

G

A

then shall

AB =DE,

AC = DF,
BACEDF.

D

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For if ABDE, then one is > other;
let AB > DE, make BG = DE,
join GC.

but

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i. e.

GB, BC = DE, EF, ea. to ea.

GBC= DEF ";

base GC base DF,

GBCDEF,

▲ GCB=/ DFE;
▲ BCA = / DFE,
▲ GCB = BCA,

less greater,

which is absurd;

.. AB is not DE,

i e.

Then

ABDE.

JAB, BC = DE, EF, ea. to each,
{≤ ABC = ≤ DEF;

Prop. 3.

Prop. 4.

Hyp.
Ax 1.

Hyp.

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AC = DF,

CASE 2. When sides oppo. equals in each
are equal to each other.
AB DE,
ZABC=_DEF,
▲ BCA = / EFD,

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then

BC= EF,

BAC / EDF,

D

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PROP. XXVI.

THEOR. 27. 1 Eu.

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 str. line EF, which falls upon the str. lines AB, CD, make alt,

EFD: then shall AB || CD.

AEF

alt.

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For if ABCD, they will, when produced, meet either towards B, D, or towards A, C. Suppose the former, and let them meet in G, then GEF will form a

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Def. 35.

;

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which is impossible.

... AB, CD, being produced, do not meet to

wards B, D.

In like manner it may be proved that they
do not meet towards A, C ;
.. AB || CD;

Wherefore if a str. line, &c.

Def. 35.

D

PROP. XXVII. THEOR. 28. 1 Eu.

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

=

Let the str. line EF, which falls upon AB
and CD, make ext. ▲ EGB = GHD, the int.
and oppo. on the same side.
Or int.s on the same side, viz.

BGH+ GHD 2 rt. Zs;
then AB || CD.

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Hyp.

Again,

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GHD=

={

ZAGH, which are alt. LS,

AB || CD.

Zs,

s,

BGH+GHD = 2 rt.

Prop. 12. and AGH+ BGH = 2 rt.

Ax. 1.

:: ≤ BGH + ≤ GHD = / AGH + ▲ BGH ;

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