Proof
Let be cut at , and produce to so that .
Then and is cut at into the segments
and . Apply II.4 to the line cut at : the square
on equals the squares on and together with twice the
rectangle on , . Since , this becomes:
\[
AD^2 \;=\; AB^2 + BC^2 + 2\cdot(AB\cdot BC).
\]
Apply II.4 again to cut at , namely , and substitute. Combining (Common Notion 2)
and re-arranging (Common Notion 3) to isolate
on the right side yields:
\[
AD^2 \;=\; 4\cdot(AB\cdot BC) + AC^2,
\]
which is in the form Euclid states it.
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Full neighborhood
Depends on (6)
- II.1Proposition II.1If there be two straight lines, and one of them be cut into any number of segments whatever, the rectangle contained by…
- 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.
- 3Common notion 3If equals be subtracted from equals, the remainders are equal.
- II.1Definition II.1Any rectangular parallelogram is said to be contained by the two straight lines containing the right angle.
- II.2Definition II.2And in any parallelogrammic area let any one whatever of the parallelograms about its diameter, with the two…
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