Proof
Let be cut at . Describe on the square (I.46),
and through draw parallel to or (I.31), meeting
at . On as side construct the square inside
(a copy of II.4's construction), with parallel to .
By II.4 the square on equals the square on plus the
square on plus twice the rectangle . Add the
square on to both sides of this identity (Common Notion 2):
\[
AB^2 + CB^2 \;=\; AC^2 + 2\cdot CB^2 + 2\cdot(AC\cdot CB).
\]
But (Definition II.1; II.1). Hence
, as required.
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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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