fquare BD would be equal to the fquare FH (II. 2.), which it is not. Neither can it be lefs, for then the fquare BD would be lefs than the fquare FH (II. 4.), which it is not; confequently AB is greater than EF, as was to be fhewn. PROP. V. THEOREM Parallelograms or triangles, having equal bases and altitudes, are equal to each other." K Let AC, EG be two parallelograms, having the base AB equal to the base EF, and the altitude DK to the altitude HL; then will the parallelogram AC be equal to the parallelogram EG. For upon AB, EF, produced if neceffary, let fall the perpendiculars CM, GN (I. 12.) Then, fince MD, NH are rectangular parallelograms, the fide DC is equal to the fide KM, and the fide HG to the fide LN (I. 30.) But DC is alfo equal to AB, and HG to EF (I. 30.); therefore KM is equal to AB, and LN to EF. And, fince AB is equal to EF (by Hyp.), KM will be equal to LN; and confequently the rectangle MD is equal to the rectangle NH (II. 2.) But the rectangle MD is equal to the parallelogram AC, because they stand upon the fame base DC, and between the fame parallels DC, AM. E 2 And, And, for the fame reafon, the rectangle NH is equal to the parallelogram EG; whence the parallelogram AC is equal to the parallelogram EGA Again, let ABD, EFH be two triangles, having the base AB equal to the bafe EF, and the altitude DK to the altitude HL; then will the triangle ABD be equal to the triangle EFH... For, if the parallelograms AC, EG be compleated, they will be equal to each other, by the former part of the propofition!... And fince the diagonals, DB, HF divide them into two equal parts (1. 30.) the triangle ABD will also be equal to the triangle EFH. Q. E. D. COROLL. Parallelograms or Triangles ftanding upon equal bafes, and between the fame parallels, are equal to each other! 1 PROP. VI. THEOREM. The complements of the parallelograms which are about the diagonal of any parallelogram are equal to each other. ( Let AC be a parallelogram, and AK, KC, complements about the diagonal BD; then will the complement AK be equal to the complement KC. For fince AC is a parallelogram, whofe diagonal is BD, the triangle DAB will be equal to the triangle BCD (I. 30.) And, because EG, HF are also parallelograms, whofe diagonals are DK, KB, the triangle DGK will be equal to the triangle DEK, and the triangle KFB to the triangle кнв (І. 30.). But, fince the triangles DGK, KFB are together, equal to the triangles DEK, KHB, and the whole triangle DAB to the whole triangle DCB, the remaining part AK will be equal to the remaining part KC. Q. E. D. Parallelograms which are about the diago nal of a fquare are themselves fquares. G с A Let BD be a fquare, and HE, FG parallelograms about its diagonal AC; then will, thofe parallelograms also be fquares. For fince the fide of the fquare AB is equal to the fide BC, the angle CAB will be equal to the angle ACB (I. 5.) And because the right line GH is parallel to the right line CB, the angle AKH will also be equal to the angle ACB (I. 25.) The angles CAB, AKH are, therefore, equal to each other; and confequently the fide AH is equal to the fide нк (І. 6.) But the fide AH is equal to the fide EK, and the fide HK to the fide AE (I. 30.); whence the figure HE is equilateral. It has alfo all its angles right angles : For EAH is a right angle, being the angle of a fquare; and HG, EF are each of them parallel to the fides of the fame fquare, whence the remaining angles will also be right angles (I. 25.) The figure HE, therefore, being equilateral, and having all its angles right angles, is a fquare; and the fame may be proved of the figure FG, Q. E. D, PROP. VIII. THEOREM. The rectangles contained under a given line and the feveral parts of another line, any how divided, are, together, equal to the rectangle of the two whole lines. Let A and BC be two right lines, one of which, BC, is `divided into several parts in the points D, E; then will the rectangle of A and BC, be equal to the fum of the rectangles of A and BD, A, and DE, and A and EC. For make BF perpendicular to BC (I. 11.) and equal to A (I. 3.), and, draw FG parallel to BC, and DH, EI and CG each parallel to BF (I. 27.) producing them till they meet FG in the points H, I, G. Then, fince the rectangle BH is contained by BD and BF (II. Def. 3.), it is alfo contained by BD and A, becaufe BF is equal to A (by Conft.) And, fince the rectangle DI is contained by DE and , DH, it is also contained by DE and A, because DH is equal to BF (I, 30.), or a. The rectangle EG, in like manner, is contained by EC and A; and the rectangle BG by BC and A. But the whole rectangle BG, is equal to the rectangles BH, DI and EG, taken together; whence the rectangle of A and BC is also equal to the rectangles of A and BD, A and DE and A and EC, taken together. Q. E. D. PROP. IX. THEOREM. If a right line be divided into any two parts, the rectangles of the whole line and each of the parts, are, together, equal to the fquare of the whole line. Let the right line AB be divided into any two parts in the point c; then will the rectangle of AB, AC, together with the rectangle of AB, BC, be equal to the fquare of AB. |