Proof
Let have an obtuse angle at , and let be the
vertex opposite a side about the obtuse angle. From drop a
perpendicular to extended through to (I.12), so
that the foot falls outside segment on the far side of .
In the right-angled triangle , Proposition I.47 gives
\[
BC^2 \;=\; BD^2 + CD^2.
\]
By the binomial-square identity II.4 applied to cut at
(with as a straight line, since lies on
extended through ):
\[
BD^2 \;=\; BA^2 + AD^2 + 2\cdot(BA \cdot AD).
\]
Substitute, and use I.47 in the right-angled triangle to
write ; then , 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 (the side
on which the perpendicular falls) and (the segment cut off
outside).
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Full neighborhood
Depends on (5)
- I.12Proposition I.12To draw a perpendicular straight line to a given infinite straight line from a given point not on it.
- I.47Proposition I.47In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides…
- II.4Proposition II.4If a straight line be cut at random, the square on the whole is equal to the squares on the segments and twice the…
- 2Common notion 2If equals be added to equals, the wholes are equal.
- 4Common notion 4Things which coincide with one another are equal to one another.
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