and therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD. [II. 12.
To each of these equals add the square on BC.
Therefore the squares on AB, BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD. [Axiom 2.
But because BD is divided into two parts at C, the rectangle DB, BC is equal to the rectangle BC, CD and the square on BC; [II. 3.
and the doubles of these are equal,
that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC.
Therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC;
that is, the square on AC alone is less than the squares on AB, BC by twice the rectangle DB, BC.
Lastly, let the side AC be perpendicular to BC.
Then BC is the straight line between the perpendicular and the acute angle at B;
and it is manifest, that the squares on AB, BC are equal to the square on AC, and twice the square on BC. [I. 47 and Ax. 2.
Wherefore, in every triangle &c. q.e.d.
PROPOSITION 14. PROBLEM.
To describe a square that shall he equal to a given rectilineal figure.
Let A be the given rectilineal figure: it is required to describe a square that shall be equal to A.
Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. [1. 45.
Then if the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done.
