one of its angular points with all the others; and successive parallelograms be applied to one another, equal to the different triangles of the polygon, just as KL was to AH in Case I. And it can be shewn by a similar proof that the resulting figure is a parallelogram fulfilling the requisites of the problem.

PROP. XLVI. PROB.

To describe a square on a given straight line.

Let AB be a given straight line. It is required to describe a square on AB.

From A draw (i. 11) AC at right angles to AB; and from ac cut off (i. 3) AD equal to AB. Through D draw (i. 31) DE parallel to AB; and through B draw BF parallel to ac, cutting DE in F. Then ABFD shall

be the square required.

The figure ABFD is a parallelo

gram by const"; and the opposite

C

D

F

E

sides of parallelograms are equal (i. 34): therefore AB is equal to DF, and AD to BF. But AB is equal to AD by const" therefore the four straight lines DA, AB, BF, FD are all equal, and the figure ABFD is equilateral. Again, because AC cuts the parallels DF, AB in D, a, the two interior angles on the same side of DA, ADF, DAB are equal (i. 29) to two right angles. But DAB is a right angle by const": therefore also ADF is a right angle; and the opposite angles of parallelograms are equal (i. 34) ; therefore each of the opposite angles ABF, BFD is a right angle: wherefore the figure ABFD is rectangular. Hence the four-sided figure ABFD has been shewn to be both equilateral and rectangular, and it is therefore a square (Def. 30), and it is described on the given straight line Which was to be done.

AB.

COR.-Every parallelogram that has one of its angles a right angle shall have all its angles right angles.

Let ABCD be a parallelogram, and let it have one of its angles DAB a right angle. Then the three other angles ABC, BCD, CDA shall also be right angles.

Because AD cuts the parallels AB, DC in AD, it may be shewn as in the prop" that the angle ADC is álso a right angle, and then that each of the other two angles opposite is a right angle. Hence ABCD is rectangular. Which was to be proved.

OBS.-It appears from the proof of this prop" that the opposite sides of a square are parallels.

In any right-angled triangle the square, which is described on the side opposite to the right angle, shall be equal to the squares, which are described on the sides including the right angle.

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

On BC, AB, AC describe (i. 46) the three squares BDEC, BFGA, AHKC; through a draw (i. 31) AL parallel to BD or CE (i. 46, Obs.), cutting DE in L; and join FC, AD.

F

B

D

G

Ꮮ

E

H

K

Because BAC is a right angle by hyps, and BAG is a right angle, since it is an angle of a square (Def. 30); therefore the two straight lines CA, GA on opposite sides of BA make with it at the point A the adjacent angles BAC, BAG equal to two right angles. Therefore CA is in the same straight line with AG. In like manner it may be shewn that BA is in the same straight line with AII. Because DBC

is a right angle, since it is an angle of a square, and

ABF is a right angle for the same reason; and all right angles are equal to one another (Ax. 11): therefore the. angle DBC is equal to the angle FBA. To each of these angles add the angle ABC; then the whole angle DBA is equal (Ax. 2) to the whole angle FBC. Also AB is equal to FB, and BD to BC, since they are sides of squares (Def. 30): therefore the two triangles DBA, FBC have the two sides AB, BD respectively equal to the two sides FB, BC and the included angle ABD equal to the included angle FBD. Therefore these two triangles are equal in every respect (i. 4), and hence the triangle ABD is equal to the triangle FBC. Now because the parallelogram (i. 46, Obs.) BL and the triangle ABD are on the same base BD and between the same parallels BD, AL; therefore the parallelogram BL is double (i. 41) of the triangle ABD. And because the square GB and the triangle CFB are on the same base FB and between the same parallels FB, GAC; therefore the square GB is double of the triangle BFC. But the triangle ABD was shewn to be equal to the triangle FBC; and the doubles of equal things are equal (Ax. 7): therefore the parallelogram BL is equal to the square GB. And in like manner by joining AE, BK it may be shewn that the parallelogram OL is equal to the square HC. Therefore, adding equals to equals, the whole square BDEC is equal to the two squares GB, HC; that is, the square described on the side BC, opposite to the right angle, is equal to the squares described on the sides AB, AC, including the right angle. Which was to be proved.

PROP. XLVIII. THEOR.

If the square described on one of the sides of a triangle be equal to the squares described on the other two sides: then the angle included by these two sides shall be a right angle.

Let the square described on вo one of the sides of the triangle ABC be equal to the squares described on the other two sides AB, AC. Then the angle BAC included by

AB, AC shall be a right angle.

B

A

E

C

From A draw (i. 11) AD at right angles to AC: from AD cut off (i. 3) AE equal to AB; and join CE. Because AE is equal to AB by const", the square of AE (i. e. the square described on AE) is equal to the square of AB. To each of these equals add the square of AC: therefore the squares of EA, AC are equal (Ax. 2) to the squares of BA, AC. But the square of EC is equal (i. 47) to the squares of EA, AC, because the angle CAE is a right angle by const"; and the square of BC is equal to the squares of BA, AC by hyps: therefore the square of EC is equal to the square of BC, and therefore EC is equal to BC. Also EA is equal to AB, and AC common to the two triangles EAC, BAC: therefore these two triangles have the three sides CE, EA, AC respectively equal to the three sides CB, BA, AC. Therefore they are equal in every respect (i. 8); and hence the angle CAE is equal to the angle CAB. But CAE is a right angle therefore also BAC is a right angle. Which was to be proved.

THE

ELEMENTS OF EUCLID.

BOOK II.

DEFINITIONS.

I.

A RECTANGLE is a parallelogram which has one of its angles a right angle (and therefore (i. 46, Cor.) all its angles right angles).

OBS. A rectangle is said to be contained by either pair of its sides which include one of the right angles.

(1) Let ABCD be a rectangle. We may either speak of it as "the rectangle contained by AB, AD, or "the rectangle contained by BA, BC," or "the rectangle contained by CB, CD," or "the rectangle contained by DA, DC."

are

D

A

C

B

Frequently the words "contained by not expressed, but understood, so that the above rectangle may be simply called either "the rectangle AB, AD," or "the rectangle BA, BC," or "the rectangle CB, CD," or "the rectangle DA, DC."

B

G

K

M

L

(2) When the rectangle contained by two straight lines is spoken of, there is not always a figure given. Thus, if the "rectangle contained by the straight lines AB, CD," or simply "the rectangle AB, CD," be spoken of, and we wish to describe it; we must take some straight line EF, and

E

C

D

H

F