In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle (namely that on which the perpendicular falls) and the straight line cut off outside by the perpendicular.
Proof
Let △ABC have an obtuse angle at A, and let C be the
vertex opposite a side AB about the obtuse angle. From C drop a
perpendicular CD to AB extended through A to D (I.12), so
that the foot D falls outside segment AB on the far side of A.
In the right-angled triangle BCD, Proposition I.47 gives
\[
BC^2 \;=\; BD^2 + CD^2.
\]
By the binomial-square identity II.4 applied to BD cut at A
(with BD=BA+AD as a straight line, since D lies on AB
extended through A):
\[
BD^2 \;=\; BA^2 + AD^2 + 2\cdot(BA \cdot AD).
\]
Substitute, and use I.47 in the right-angled triangle ACD to
write AC2=AD2+CD2; then AD2+CD2=AC2, and
substitution gives:
\[
BC^2 \;=\; BA^2 + AC^2 + 2\cdot(BA \cdot AD),
\]
which is the law of cosines as Euclid states it: the square on the
side subtending the obtuse angle exceeds the sum of the squares on
the sides containing it by twice the rectangle on BA (the side
on which the perpendicular falls) and AD (the segment cut off
outside).