BD is equal to twice the rectangle of ab, bd, together And if, to each of these equals, there be added the But the sum of the squares of BD, DC is equal to the square of BC, and the fun of the squares of AD, DC to the square of ac (II. 14.); whence the square of ac is less than the fun of the squares of AB, BC, by twice the rectangle of AB, BD. QED PRO P. XIX. THEORE M. In any triangle, the double of the square of a line drawn from the vertex to the middle of the base, together with double the square of the semi-base, is equal to the sum of the squares of the other two sides. Let asc be a triangle, and ce a line drawn from the vertex to the middle of the base AB; then will twice the fum of the squares of CE, EA be equal to the sum of the squares of AC, cp. For on AB, produced if necessary, let fall the perpendicular CD (I.12.) F 2 1 2 Then, Then, because aec is an obtuse angle, the square of Ac is equal to the squares of AE, EC together with twice the rectangle of AE, ED (II. 17.) And, because Bec is an acute angle, the square of CB together with twice the rectangle of BE, Ed is equal to the squares of BE, EC (II. 18.) And since Ae is equal to EB (by Conft.), the square of BC together with twice the rectangle of AE, ED is equal to the squares of AE, EC. But if equals be added to equals, the wholes will be equal ; whence the squares of AC, CB, together with twice the rectangle of AE, ED, are equal to twice the squares of AE, EC, together with twice the rectangle of AE, ED. And, if twice the rectangle of AE, EC, which is common, be taken away, the sum of the squares of AC, CB will be equal to twice the sum of the squares of AE, EC. Q. E. D. PROP. XX. THEOREM. In an isosceles triangle, the square of a line drawn from the vertex to any point in the base, together with the rectangle of the segments of the base, is equal to the square of one of the equal fides of the triangle. Let ABC be an isosceles triangle, and ce a line drawn from the vertex to any point in the base AB; then will the {quare of ce, together with the rectangle of AE, EB be equal to the square of ac or CB. For bifect the base al in D (I. 10.), and join the points E, D. Then, fince Ac is equal to CB, AD to DB, and cd is common to each of the triangles ACD, BCD, the angle CdA will be equal to the angle CDB (I. 7.); and consequently cd will be perpendicular to AB (Def. 8, 9.) And, because ace is a triangle, and cp is the perpendicular, the difference of the squares of ac, CE is equal to the difference of the squares of AD, DE (II. 16.) But, fince be is the sum of AD and De, and ae is their difference, the difference of the squares of AD, DE is equal to the reclangle of AE, EB; confequently, the difference of the squares of ac, ce is also equal to the rect. angle AE, EB. And if, to each of these equals, there be added the square of ce, the square of ac will be equal to the square of ce, together with the rectangle of AE, EB. Q. E. D. : F 3 PROP. PRO P. XXI. THEOREM. The diagonals of any parallelogram bisect each other, and the sum of their squares is equal to the sum of the squares of the four sides of the parallelogram. Let ABCD be a parallelograin, whose diagonals AC, BD intersect each other in E; then will AE be equal to Ec, and be to ED; and the sum of the squares of ac, -BD will be equal to the sum of the squares of AB, BC, CD and DA. For fince AB, DC are parallel, and ac, BD intersect them, the angle pce will be equal to the angle EAB (I. 24.), and the angle cde to the angle EBA (I. 24.) The angle dec is likewise equal to the angle AEB (I. 15.), and the fide dc to the side AB (I. 30.); con. sequently de is also equal to es, and ce to EA (I. 21.) Again, fince DB is bifected in E, the sum of the squares of Dc, co will be equal to twice the sum of the squares of DE, EC (II. 19.) And, because pc is equal to ab, and cb to'da (I. 30.) the fum of the squares of AB, CB, DC and DA așe equal to four times the sum of the squares of DE, EC. But four times the square of de is equal to the square of BD (II. 11. Cor.), and four times the square of Ec is equal to the square of AC; whence the sum of the squares of AC, BD are equal to the sum of the squares of AB, BC, cp and DA. Q. E. D. PRO P. XXII. PROBLEM To divide a given right line into two parts, so that the rectangle contained by the whole line and one of the parts, shall be equal to the square of the other part. B H E Let all be the given right line; it is required to divide it into two parts, so that the rectangle of the whole line and one of the parts shall be equal to the square of the other part. Upon AB describe the square AC (II. 1.), and bifect the side of it Ad in E (I. 10.) Join the points B, E; and, in EA produced, take EF equal to EB (I. 3.); and upon af describe the square FH (II. 1.) Then will AB be divided in h so, that the rectangle AB, BH, will be equal to the square of AH. For, since DF is equal to the sum of EB and Ed, or its equal EA, and as is equal to their difference, the rect. angle of DF, FA is equal to the difference of the squares of EB, EA (II. 13.) |