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Right Line B C, and the Squares GB, HC, on B A,
AC. Therefore the Square B E, described on the
Side B C, is equal to the Squares described on the sides
BA, AC. Wherefore, in any Right-angled Triangle, the
Square described upon the Side, subtending the Right Angle,
is equal to both the Squares described upon the Sides, con-
taining the Right Angle ; which was to be demonstrated.
PROPOSITION XLVIII.

THE OR E M.
If a Square described upon one Side of a Triangle,

be equal to the Squares described upon the other
two Sides of the said Triangle ; then, the Angle
contained by those two other Sides is a Right

Angle. If the Square, described upon the Side B C of the

Triangle A B C, be equal to the Squares described upon the other two sides of the Triangle B A, AC; I say the Angle BAC is a Right one.

For, let there be drawn A D from the Point A, at Right Angles, co A C: Likewise make A D equal to B A, and join D C..

Then, because D A is equal to AB, the Square defcribed on D A will be equal to the Square described on AB.' And adding the common Square described on A C, the Squares described on DA, A C, are equal to the Squares described on BA, AC. But the Square described on D.C is * equal to the Squares described • 47 of ibis. on DA, A C; for DAC is a Right Angle: But the Square on B C is put equal to the Squares on B A, AC. Therefore the Square described on D C is equal to the Square described on B C; and so the Side CD is equal to the Side Ç B. And because D A is equal to A B, and A C is common, the two Sides DA, AC, are equal to the two Sides BA, AC; and the Base DC is equal to the Base CB. Therefore the Angle DAC is I equal to the Angle B A C; but D AC is 1 8 of ibis. a Right Angle; and so B A C will be a Right Angle also. If, therefore, a Square described upon one Side of a Triangle, be equal to the Squares, described upon the other two Sides of the said Triangle, then the Angle contained by these two other Sides, is a Right Angle; which was to be demonstrated.

EUCLI D's

E LE MEN T S.

BOOK II.

DEFINITION S.

1. EVERY Right-angled Parallelogram is

said to be contained under two Right Lines, comprehending a Right Angle.

II. In every Parallelogram, either of those Pa

rallelograms, that are about the Diameter, together with the Complements, is called a Gno

mon.

PRO

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THEOREM.
If there be two Right Lines, and one of them be

divided into any Number of Parts; the Reet-
angle comprehended under the whole Line and.
the divided Line, shall be equal to all the Real-
angles contained under the whole Line, and the
Several Segments of the divided Line.

L

ET A and B C be two Right Lines, whereof
B C is cut or divided any how in the Points

D, E. I say, the Rectangle contained under the Right Lines A and B C, is equal to the Rectangles contained under A and B D, A and D E, and A and E C.

For, let * BF be drawn from the Point B, at * 11. 1. Right Angles, to BC; and make + B G equal to A; +3. and let IGH be drawn through G parallel to BC:1 31. 1. Likewise let I there be drawn D C, EL, CH, thro' D, E, C, parallel to B G.

Then the Rectangle B H, is equal to the Rectangles BK, DL, and E H ; but the Rectangle B H is that contained under A and BC; for it is contained under GB, BC; and G B is equal to A; and the Rectangle BK is that contained under A and B D; for it is contained under G B and B D; and G B is equal to A; and the Rectangle D L is that contained under A and D Е, because D K, that is, BG, is equal to A : So likewise the

Rectangle E H is that contained under A and E C. Therefore the Rectangle under A and BC is equal to the Rectangles under A and B D, A and D E, and A and E C. Therefore, if there be two Right Lines given, and one of them be divided into any Number of Parts, the Rectangle comprehended under the whole Line and the divided Line, fhall be equal to all the Rectangles contained under the whole Line, and the several Segments of the divided Line ; which was to be demonftrated.

PRO

PROPOSITION II.

46. So

THEO R E M.
If a Right Line be any how divided, the Rectan-

gles contained under the wbole Line, and each
of the Segments, or Paris, are equal to the

Square of tbe whole Line.
L

E T the Right Line A B be any how divided in the

Point C. I say, the Rectangle contained under A B and BC, together with that contained under A B, and A C, is equal to the Square made on A B.

For let the Square A D E B be described * on A B, and thro' C let C F be drawn parallel to A D or BE. Therefore A E is equal to the Rectangles A F and CE. But A E is a Square described upon A B ; and A F is the Rectangle contained under B A and A C; for it is contained under D A and A C, whereof A D is equal to A B; and the Rectangle CE is contained under A B and B C, since B E is equal to A B. Wherefore the Rectangle under A B and A C, together with the Rectangle under A B and BC, is equal to the Square of A B. Therefore, if a Right Line be any how divided, the Reclangles contained under the whole Line, and each of the Segments, or Parts, are equal to the Square of the whole Line ; which was to be demonstrated.

PROPOSITION III.

THEOREM.
If a Right Line be any bow cut, ibe ReElangle

contained under the whole Line, and one of its
Parts, is equal to the Rectangle contained un-
der the two Parts, togetber with the Square of

the first-mentioned Part. LET the Right Line A B be any how cut in the

Point C. I say, the Rectangle under A B and B C is equal to the Rectangle under AC and B C, together with the Square described on B C,

For

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