Proof
Let be acute-angled, with the acute angle at .
From drop a perpendicular to (I.12). Since the angle
at is acute, the foot falls within the segment , between
and .
Apply Proposition II.7 to cut at : . In the right-angled triangle , I.47
gives , and in likewise
. Subtracting (Common Notion 3) the first
from the third: . Combining with II.7
and rearranging (Common Notion 3 + Common Notion 2):
\[
AC^2 \;=\; AB^2 + BC^2 - 2\cdot(AB \cdot BD),
\]
which is the acute-angle form of the law of cosines: the square on
the side subtending the acute angle is less than the sum of the
squares on the sides containing it by twice the rectangle on
and (the segment cut off within).
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Full neighborhood
Depends on (7)
- 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.7Proposition II.7If a straight line be cut at random, the square on the whole and that on one of the segments both together are equal to…
- II.12Proposition II.12In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides…
- 2Common notion 2If equals be added to equals, the wholes are equal.
- 3Common notion 3If equals be subtracted from equals, the remainders are equal.
- 4Common notion 4Things which coincide with one another are equal to one another.
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