II.13 Proposition II.13
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle (namely that on which the perpendicular falls) and the straight line cut off within by the perpendicular.
Proof
Let △ABC be acute-angled, with the acute angle at B.
From C drop a perpendicular CD to AB (I.12). Since the angle
at B is acute, the foot D falls within the segment AB, between
A and B.
Apply Proposition II.7 to AB cut at D: AB2+DB2=2⋅(AB⋅DB)+AD2. In the right-angled triangle BCD, I.47
gives BC2=BD2+CD2, and in △ACD likewise
AC2=AD2+CD2. Subtracting (Common Notion 3) the first
from the third: AC2−BC2=AD2−BD2. 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 AB
and BD (the segment cut off within).
lines 74–74 in main.tex
