BOOK I. PROP. XXVI. THEOR.
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and therefore =
which is absurd;
hence neither of the sides
and
is greater than the other; and ∴ they are equal;
CASE II.
Again, let =
which lie opposite
the equal angles
and
. If it be possible, let
>
, then take
=
, draw
.
Then in
and
we have
=
,
∴ =
which is absurd (pr. 16.).
Consequently, neither of the sides or
is greater than the other, hence they must be equal. It follows (by pr. 4.) that the triangles are equal in all respects.
Q. E. D.