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PROPOSITION 47. THEOREM. In any right-angled triangle, the square which is de scribed on the side subtending the right angle is equal to the squares described on the sides which contain the right angle.

Let ABC be a right-angled triangle, having the right angle BAC: the square described on the side BC shall be equal to the squares described on the sides BA, AC.

. On BC describe the square BDEC, and on BA, AC describe the squares GB, HC; II. 46. through A draw AL parallel to BD or CE;

[I. 31. and join AD, FC.

Then, because the angle BAC is a right angle, (Hypothesis. and that the angle BAG is also a right angle, [Definition 30. the two straight lines AC, AG, on the opposite sides of AB, make with it at the point À the adjacent angles equal to two right angles ; therefore CA is in the same straight line with AG. [I. 14. For the same reason, AB and AH are in the same straight

line.

Now the angle DBC is equal to the angle FBA, for each of them is a right angle.

[Axiom 11. Add to each the angle ABC. Therefore the whole angle DBA is equal to the whole angle FBC.

. [Axiom 2. And because the two sides AB, BD are equal to the two sides FB, BC, each to each ;

[Definition 30. and the angle DBA is equal to the angle FBC; therefore the triangle ABD is equal to the triangle FBO.

[I. 4.

Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL.

[1. 41. And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC.

[I. 41. But the doubles of equals are equal to one another. [Ax. 6. Therefore the parallelogram BL is equal to the square GB.

In the same manner, by joining AE, BK, it can be shewn, that the parallelogram (L is equal to the square CH. Therefore the whole square BDEC is equal to the two squares GB, HC.

[Axiom 2. And the square BDEC is described on BC, and the squares GB, HC on BA, AC. Therefore the square described on the side BC is equal to the squares described on the sides BA, AC.

Wherefore, in any right-angled triangle &c. Q.E.D.

PROPOSITION 48. THEOREM. If the square described on one of the sides of a triangle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle.

Let the square described on BC, one of the sides of the triangle ABC, be equal to the squares described on the other sides , AC: the angle BAC shall be a right angle.

From the point A draw AD at right angles to AC; [I. 11. and make AD equal to BA; '[I. 3. and join DC.

Then because DA is equal to BA, the square on DA is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA, AC.

[Axiom 2.

But because the angle DAC is a right angle, [Construction. the square on DC is equal to the squares on DA, AC.[I. 47. And, by hypothesis, the square on BC' is equal to the squares on BA, AC. Therefore the squareon DC is equal to the square on BC![Ax. 1. Therefore also the side DC is equal to the side BC.

And because the side DA is equal to the side AB;

[Constr. and the side AC is common to the two triangles DAC, BAC; the two sides DA, AC are equal to the two sides BA, AC, each to each ; and the base DC has been shewn to ! be equal to the base BC; therefore the angle DAC is equal to the angle BAC. [I. 8. But DAC is a right angle;

[Construction. therefore also BAC is a right angle.

[Axiom 1. Wherefore, if the square &c. Q.E.D.

BOOK II.

DEFINITIONS.

1. EVERY right-angled parallelogram, or rectangle, is said to be contained by any two of the straight lines which contain one of the right angles.

2. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon.

Thus the parallelogram HG, Ą_ together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelo 1 7 grams which make the gnomon.

PROPOSITION 1. THEOREM.

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

Let A and BC be two straight lines ; and let BC be divided into any number of parts at the points D, E: the rectangle contained by the straight lines A, BC, shall be equal to the rectangle contained by A, BD, together with that contained by X, DE, and that contained by A, EC.

From the point B draw BF at right angles to BC; [I. 11.

B

__ D E_ and make BG equal to A ; [I. 3. through G draw GH parallel to BC; and through D, E, C draw DK, EL, CH, parallel to ᏴG.

[I. 31.

K Then the rectangle BH is equal to the rectangles BK, DL, EH. But BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A.

[Construction. And BK is contained by A, BD, for it is contained by GB, BD, and GB is equal to A; and DL is contained by A, DE, because DK is equal to BG, which is equal to A;

[I. 34. and in like manner EH is contained by A, EC. Therefore the rectangle contained by A, BC is equal to the rectangles contained by A, BD, and by A, DE, and by A, EC.

Wherefore, if there be two straight lines &c. Q.E.D.

PROPOSITION 2. THEOREM.

If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square on the whole line.

[1.46.

Let the straight line AB be divided into any two parts at the point C: the rectangle contained by AB, BC, together with the rectangle AB, AC, shall be equal to the square on AB.

[Note. To avoid repeating the word А с В contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC.]

On AB describe the square ADEB; and through C draw CF parallel DF E to AD or BE.

[1.31. Then AE is equal to the rectangles AF, CE. But AE is the square on AB. And AF is the rectangle contained by BA, AC, for it is contained by DA, AC, of which DA is equal to BA; and CE is contained by AB, BC, for BE is equal to AB. Therefore the rectangle AB, AC, together with the rectangle AB, BC, is equal to the square on AB.

Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 3. THEOREM.

If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part.

Let the straight line AB be divided into any two parts at the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square on BC

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