duced to L; and produce FA to H. In the parallelogram EDKL thus constructed, the parallelograms LA, AD are the complements of those about the diagonal, and are therefore equal (Prop. 43); but the parallelogram AD is equal to the given triangle G, therefore LA also is equal to G; and it is applied to the given line AB: and also since ACD =▲▼ (Const.), and /ACD=/FAB=LAHL (Prop. 29), the parallelogram AL has an angle AHL equal to the given angle V. PROP. XLV. PROB. To construct a parallelogram equal to a given rectilinear figure (ABCDE) and having an angle equal to a given rectilinear angle (N). D G H C K E L M N P B Draw AC and EC dividing the rectilinear figure into triangles, and construct the parallelogram GK equal to the triangle EDC, and having the angle GIK equal to the given angle N (Prop. 42). To the line IK apply the parallelogram IM equal to the triangle ECA, and having the angle A O ILM equal to the angle N (Prop. 44); and to the line LM in like manner apply the parallelogram LP equal to the triangle ACB and having the angle LOP equal to the angle N. Now, since s GIK, ILM are both equal to ≤ N (Const.), they are equal to each other (Ax. 1); and 2 KIL being added to both, then / GIK + KIL = / KIL + ▲ ILM; but the latter are equal to two right angles (Prop. 29); therefore Zs GIK, KIL are also equal to two right angles, and the lines GI, IL must form one continued straight line (Prop. 14). And in like manner it may be shown that LO is in the same straight line with IL, and, consequently, that GO is one straight line. And since GO and HK are parallel, the alternate angles HKI, KLI are equal (Prop. 29), and therefore IKM being added to both, HKI+ZIKM=/KIL+ZIKM; but the latter are equal to two right angles (Prop. 29); therefore Zs HKI, IKM are also equal to two right angles, and the lines HK and KM form one straight line (Prop. 14). And in like manner it may be shown that MP forms one straight line with KM, and therefore HP is one straight line, and it is parallel to GO. And also, since GH is parallel to IK, and IK to LM, then GH is parallel to LM (Prop. 30); and since LM is parallel to OP, GH is also parallel to OP. Since, therefore, HP is parallel to GO, and GH to OP, the figure GOPH is a parallelogram; and since its parts, GK, IM, LP are severally equal to the triangles EDC, ECA, ACB, the parallelogram GOPH, is equal to the rectilinear figure ABCDE; and it has an angle GOP equal to the given angle N (Const.). PROP. XLVI. PROB. Upon a given straight line (AB) to construct a square. D From the point A draw AC perpendicular to AB (Prop. 11), and make it equal to AB (Prop. 3). Through Cc draw CD parallel to AB, and meeting BD drawn from B parallel to AC. In the parallelogram ACDB the side AC=AB (Const.); and since the angle A is a right angle (Const.), the angle B must also be a right angle (Prop. 29), and the remaining A sides and angles must be equal (Prop. 34); and therefore the parallelogram ACDB having its sides and angles equal is a square (Def. 27). B In a right angled triangle (ACB) the square of the hypotenuse (AB), or side subtending the right angle, is equal to the squares of the sides (AC, BC) which contain the right angle. On AB construct the square ADEB, and on AC and BC construct the squares AP and BR. From C draw CF parallel to AD or BE, and join CD and BG. Then, since Zs ACB, ACP are both right angles, the lines BC and CP form one straight line (Prop. 14), and in like manner it may be shown that AC and CR also form one straight line. And since s DAB, GAC are both right angles, and therefore equal (Ax. 10), if BAC be added to both, then DAC Z BAG (Ax. 2), and also the sides containing those angles are respectively equal, viz. DA=BA and AC=AG. Therefore the triangle DAC=BAG (Prop. 4): but the parallelogram = G Р B R FA is double of the triangle DAC, because they are on the same base AD, and between the same parallels AD, CF (Prop. 41); and in like manner the square AP is double of the triangle BAG, because they are on the same base GA, and between the same parallels GA, PB; therefore the parallelogram AF is equal to the square AP. And in like manner it may be shown that the parallelogram BF is equal to the square BR. The two parallelograms AF and BF are therefore equal to the squares AP and BR; but the parallelograms AF and BF together form ADEB, the square of the side AB subtending the right angle, which is therefore equal to the squares AP and BR, which are the squares of the sides AC and CB containing the right angle. D F E If the square of one side (AB) of a triangle is equal to the squares of the other two sides (AC, CB), the angle (ACB) subtended by that side is a right angle. B From the point C draw CD perpendicular to BC, and make it equal to CA and join DB. Then, since CD and CA are equal, their squares are equal, and if the square of CB be added to both, the squares of DC and CB are together equal to the squares of AC and CB; but the square of DB is equal to the squares of DC A and CB (Prop. 47), and the square of AB is equal to the squares of BA and AC (Hyp.). The square of DB therefore is equal to the square of AB (Ax. 1), and therefore DB=AB. Since therefore the triangles ACB, DCB have their sides and bases equal, they shall have the angles contained by the equal sides also equal (Prop. 8), and therefore / ACB=/ DCB, that is to say, the angle ACB is a right angle. Note. For Remarks and Exercises, see APPENDix B. BOOK II. DEFINITIONS. D C 1. A RECTANGLE or right-angled parallelogram is said to be contained by any two of its adjacent sides. [Thus the right-angled parallelogram ADCB, or AC, is said to be contained by the sides AB and BC; or it may be briefly designated the rectangle AB BC. If the adjacent sides AB and BC are equal, then AB BC is A a square, and is equal to AB AB, or, as it may cisely written, AB2.] C B be con 2. In a parallelogram, the figure composed of one of the parallelograms about the diagonal, together with the two complements, is called a GNOMON. [Thus the parallelogram EK, which is about the diagonal, forms, with the com-plements CF, FB, the gnomon HAG; and the parallelogram HG, with the complements, forms the gnomon EDK.] E H D/ G F A K B PROP. I. THEOR. The rectangle contained by two straight lines (L and AB), of which one (AB) is divided into any number of parts (AF, FH, HB), is equal to the sum of the rectangles contained by the undivided line and the several parts of the divided line. From A draw AC perpendicular to AB, make it equal to L, and complete the parallelogram AD (that is to say, draw D E G D F H B CD parallel to AB and meeting BD drawn parallel to AC); and from the points F, H draw FE, HG C parallel to AC. Then AC AB is a rectangle (Const.), and it is divided into the rectangles AC AF, FE FH, HG HB, to the sum of which it is equal. But since AC=L (Const.), A the rectangle AC·AB=L'AB; and in like manner AC AF=L'AF; and also, because FE, HG are equal to AC (i. Prop. 34), FE·FH=L·FH, and HG⚫HB=L•HB. Therefore, L'AB, or the rectangle contained by the two lines, is equal to L'AF+L FH+L HB, or the sum of the rectangles contained by the undivided line and the parts of the divided line. L PROP. II. THEOR. If a straight line (AB) be divided into any two parts (AF, FB), the square of the whole line is equal to the sum of the rectangles contained by the whole line and each of its parts. C E D On AB describe the square ACDB (i. Prop. 46), and from F draw FE parallel to AC or BD. Then ACDB or AB AC AF+BD.BF. But since AC=AB =BD, the rectangle AC AF = AB AF, and BD BF AB BF. Therefore AB', or the square of the whole line, is equal to AB·AF + AB·BF, or the sum of the rectangles contained by the whole line and each of its parts. A F B If a straight line (AB) be divided into any two parts (AF, FB), the rectangle contained by the whole line and either of its parts, is equal to the square of that part, together with the rectangle under the parts. Upon either of the parts, as AF, describe the square ACEF, and complete the parallelogram FD, by producing CE till it meets BD drawn parallel to FE or AC. Then |