Then, since he is a rectangle, contained by AB, BE, it is also contained by AB, BC, because we is equal to BC (II. Def. 2.) And, in like manner, Ad is a rectangle contained by AC, CD, or by AC, CB; and ce is the square of CB (by Conft.) But the rectangle Ae is equal to the rectangle AD, and the square ce, taken together ; whence the rectangle of AB, BC is also equal to the rectangle of AC, CB together with the square of CB. R. E. D. PROP. XI. THEOREM. If a right line be divided into any two parts, the square of the whole line will be equal to the squares of the two parts, together with twice the rectangle of those parts. Let the right line AB be divided into any two parts in the point c; then will the square of AB be equal to the squares of AC, CB together with twice the rectangle of AC, CB. For upon AB make the square AD (II. 1.), and draw the diagonal EB; and make ck, FH parallel to AE, ED (I. 27.): Then, Then, since the parallelograms about the diagonal of a square are themselves squares (II. 7.), FK will be the {quare of FG, or its equal ac, and ch of CB, And since the complements of the parallelograms about the diagonal are equal to each other (II. 6.), the complement AG will be equal to the complement GD. But AG is equal to the rectangle of AC, CB, because cg is equal to CB (II. Def. 2.); and GD is also equal to the rectangle of ĄC, CB, because GK is equal to GF (Def. II. 2.) or AC (I. 30.), and GH to CB (I. 30.) The two rectangles AG, GD are, therefore, equal to twice the rectangle of AC, CB; and FK, CH have been proved to be equal to the squares of AC, CB. But these two rectangles, together with the two squares, make up the whole square AD; consequently the square AD is equal to the squares of AC, CB, together with twice the rectangle of AC, CB. . E. D. COROLL. If a line be divided into two equal parts, the square of the whole line will be equal to four times the square of half the line. PRO P. XII. THEOREM. If a right line be divided into any two parts, the squares of the whole line, and one of the parts, are equal to twice the rectangle of the whole line and that part, together with the square of the other part. Let the right line AB be divided into any two parts in the point c; then will the squares of AB, BC, be equal to twice the rectangle AB, BC together with the square of AC. For, upon AB make the square AD (II. 1.), and draw the diagonal BE; and make FC, HK parallel to BD, BA (I. 27.): Then because AG is equal to GD (II. 6.), to each of these equals add ck, and the whole AK will be equal to the whole cd. And, since the doubles of equals are equal, the gnomon HBF, together with ck, will be the double of AK. But ck is a square upon cB (II. 7.), and twice the rectangle AB, BC is the double of AK, whence the gnomon HBF, together with the square ck, is, also, equal to twice the rectangle AB, BC, And, And, because hy is a square upon hs or ac (II. 7.), if this be added to each of these equals, the gnomon HBF, together with the squares CK, HF, will be equal to twice the rectangle AB, BC, together with the square of Ac. But the gnomon HBF, together with the squares cK, HF, are equal to the whole square AD, together with the square ck; consequently, the squares of AB, BC, are equal to twice the rectangle AB, BC together with the square of Ac. Q. E, D. PRO P. XIII. THEOREM. The difference of the squares of any two unequal lines, is equal to a rectangle under their sum and difference, Let AB, AC be any two unequal lines ; then will the difference of the squares of those lines be equal to a rectangle under their sum and difference. For, upon AB, AC make the squares AE, AI (II. 1.); and in He, produced, take EG equal to Ac (I. 3.); and make GF parallel to EB (I. 27.); and produce ci, IK till they meet HG, GF in D and F. Then, since he is equal to AB (Def. II. 2.) and Eg to AC (by Conft.), HG will be equal to the sum of AB and Aca And because Ah is equal to AB, and AK to AC (II. Def. 2.), KH will be equal to cb, or the difference of AB and Ac. But the rectangle KG is contained by HG, and HK, whence it is, also, contained by the sum and difference of aB and Ac. And, since le is equal to HK (I. 30.) or CB (by Conf.), and Eg to Ac (by Conft.) ci, or LB, the rectangle LG will be equal to the rectangle Lc (II. 2.) But the rectangles Ht, Lc are, together, equal to the difference of the squares AE, AI; consequently the rectangles hi, LG, or the whole rectangle KG, is also equal to the difference of those squares. Q. E. D. 1 PRO P. XIV. THEOREM. In any right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Let ABC be a right angled triangle, having the right angle ACB; then will the square of the hypotenuse AB he equal to the sum of the squares of ac and CB. For, on AB, describe the square AE (II. 1.), and on AC, CB the squares AG, BH; and, through the point c, dray |