The celebrated proposition of Pythagoras (1. 35) may be established as follows.

Let ABC be a right-angled s having the right - ACB.

On AC, CB describe squares XC, YC. Then ACE, BCD are straight lines equal and I to one another.

Complete the square XY.

Draw AS, BT 1 to AB, and join ST.
Then as AXS, TYB are each= A ACB in all respects;

AS is = and || to BT,
Hence BS is a o, and .. the square on AB.

Now, square on AB + four as each equal to A ABC = square XY;

also squares on AC, CB + four as each equal to A ABC = square XY;

... square on AB = squares on AC, CB.

BOOK II.

RECTANGLES.

PROPOSITION I. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line and the several parts of the divided line.

Let A and BC be two straight lines,
and let BC be divided into any parts BP, PQ, QC.

Then shall the rectangle A, BC be = rectangle A, BP + rectangle A, PQ + rectangle A, QC.

COR. 1. If A be = BC, then the rectangle (A, BC) is = the square on BC.

.. square on BC is = rectangle BC, BP + rectangle BC, PQ + rectangle BC, QC.

Hence, if a straight line be divided into any number of parts, the square on the whole line is equal to the sum of the rectangles contained by the whole line and the several parts.

COR. 2. If A be = one of the parts of the divided line, as BP,

then the rectangle A, BP is = the square on BP.

.: rectangle BP, BC = square on BP + rectangle BP, PQ + rectangle BP, QC.

Hence, if a straight line be divided into any number of parts, the rectangle contained by the whole line and one of the parts is equal to the square on that part together with the rectangles contained by that part and the other parts. "

PROPOSITION II.

If A and B be two straight lines and S their sum, the square on. S is > the squares on A and B by twice the rectangle A, B. Also, if A and B be unequal and D be their difference, the rectangle S, D is = the difference between the squares on A, B; and the square on D is < the squares on A and B by twice the rectangle A, B.

Since S is = A together with B; .. rectangle S, A = square on A + rectangle A, B, (11. 1).

rectangle S, 'B = rectangle A, B + square on B, (11. I ) and square on S = rectangle S, A + rectangle S, B, (11. 1) .:: square on S = squares on A, B + twice rectangle A, B.

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Again, since D + B = A; .. rectangle S, D + rectangle S, B = rectangle S, A, (11.1) .. rectangle S, D + rectangle A, B + square on B

= square on A + rectangle A, B ; .. rectangle S, D + square on B = square on A.

Cor. It follows from the proposition that squares on S, D + twice rectangle A, B

= twice squares on A, B + twice rectangle A, B; .. squares on S, D= twice squares on A, B,

Also that square on S= squares on A, B + twice rectangle A, B,

and .:. = square on D + four times rectangle A, B.

N.B. By the method here adopted all propositions which are the Geometrical interpretations of Algebraical identities of the second degree may be established.