(Edited by Sir Thomas L. Heath, 1908)

[Euclid, ed. Heath, 1908, on

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

Let ABC,
DEF be two triangles
having the two angles ABC,
BCA equal to the two
angles DEF,
EFD respectively, namely
the angle ABC to the
angle DEF, and the angle
BCA to the angle
EFD; and let them also have
one side equal to one side, first that adjoining the
equal angles, namely BC to
EF;

I say that they will also have the remaining
sides equal to the remaining sides respectively,
namely AB to
DE and
AC to
DF, and the remaining
angle to the remaining angle, namely the angle
BAC to the angle
EDF.

For if AB is unequal to DE, one of them is greater.

Let AB be greater, and let
BG be made equal to
DE;

and let GC be joined.

Then, since BG
is equal to DE,
and BC to
EF,

the two sides GB,
BC are equal to the two sides
DE, EF
respectively;

and the angle GBC is equal to
the angle DEF;

therefore the base GC is equal
to the base DF,

and the triangle GBC is equal
to the triangle DEF,

and the remaining angles will be equal to the
remaining angles, namely those which the equal sides
subtend;
[I. 4]

therefore the angle GCB is
equal to the angle DFE.

But the angle DFE
is by hypothesis equal to the angle
BCA;

therefore the angle BCG is
equal to the angle BCA,

the less to the greater: which is impossible.

Therefore AB is not unequal
to DE,

and is therefore equal to it.

But BC is also equal to
EF;

therefore the two sides AB,
BC are equal to the
two sides DE,
EF respectively,

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.
[I. 4]

Again, let sides subtending equal angles be
equal, as AB to
DE;

I say again that the remaining sides will
be equal to the remaining sides, namely
AC to DF
and BC to
EF, and further the remaining angle
BAC is equal to the remaining
angle EDF.

For if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; let AH be joined.

Then, since BH
is equal to EF,
and AB to
DE,

the two sides AB,
BH are equal to the two sides
DE, EF
respectively, and they contain equal angles;

therefore the base AH is equal
to the base DF,

and the triangle ABH is equal
to the triangle DEF,

and the remaining angles will be equal to the
remaining angles, namely those which equal sides
subtend;
[I. 4]

therefore the angle BHA is
equal to the angle EFD.

But the angle EFD is equal to
the angle BCA;

therefore, in the triangle AHC,
the exterior angle BHA
is equal to the interior and opposite angle
BCA:

which is impossible.
[I. 16]

Therefore BC is not unequal
to EF,

and is therefore equal to it.

But AB is also equal to
DE;

therefore the two sides AB,
BC are equal to the
two sides DE,
EF respectively, and they
contain equal angles;

therefore the base AC is equal
to the base DF,

the triangle ABC equal to the
triangle DEF,

and the remaining angle BAC
equal to the remaining angle
EDF.
[I. 4]

Therefore etc. Q.E.D.

Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)

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