Let the rhomboids BD and FH have the angle BAD equal to FEH and the containing sides AB and AD equal respectively to FE and EH; these rhomboids are equal. For if the rhomboid BD be applied to FH, the angle BAD will adapt to FEH, and its sides being equal, the points B and D must coincide with F and H, consequently the diagonal A BD will coincide with FH. Whence the two triangles BCD and FGH, having all their sides respectively equal, must also fit with each other. Cor. Hence the squares of equal straight lines are equal, and conversely equal squares have equal bases. PROP. X. THEOR. The complements of the rhomboids about the diagonal of a rhomboid, are equivalent. Let EI and HG be rhomboids about the diagonal of the rhomboid BD; their complements BF and FD contain equal spaces. A B H C G Since the diagonal AF bisects the rhomboid EI (I. 29. cor.), the triangle AEF is equivalent to AIF; and for the same reason, the triangle FHC is equivalent to FGC. From the whole triangle ABC on the one side of the diagonal, take away the two triangles AEF and FHC; and from the triangle ADC, which is equal to it, take away, on the other side, the two triangles AFI and FGC, and there remains the rhomboid BE equivalent to FD. PROP. XI. PROB. With a given straight line to construct a rhom boid equivalent to a given rectilineal figure, and having its angle equal to a given angle. Let it be required to construct, with the straight line L, a rhomboid, containing a given space, and having an angle equal to K. Construct (II. 7.) the rhomboid BF equivalent to the rectilineal figure, and having an angle BEF equal to K; produce EF until FG be equal to L, through G draw DGC parallel to EB and meeting the production of BH in C, join CF and produce it to meet the production of BE in A; draw AD parallel to EF, meet B H E ing CG in D, and produce HF to I: FD is the rhomboid required. For FD and FB are evidently complementary rhomboids, and therefore (II. 9.) equivalent; and by reason of the parallels AE, IF, the angle FID is equal to EAI (I. 34.), which again is equal to BEF or the given angle K. PROP. XII. THEOR. A trapezium is equivalent to the rectangle contained by its altitude and the part of the base cut off by a perpendicular from its remoter summit. Let ABCD be a trapezium, and CE a perpendicular drawn from C to the base AD; the trapezium is equal to the rectangle contained by AE and CE. For complete the rectangle EF. The triangles ABF and CDE have, from the definition of the trapezium, the side AB equal to CD, AF to CE, and the right angle AFB equal to CED; where F B A ED fore these triangles, being also of the same affection, are D equal (I. 24.). To each of them, add the quadrilateral space ABCE, and the rectangle AFCE is equal to the tra pezium ABCD. PROP. XIII. THEOR. A trapezoid is equivalent to the rectangle contained by its altitude and half the sum of its parallel sides. The trapezoid ABCD is equivalent to the rectangle contained by its altitude and half the sum of the parallel sides BC and AD. For draw CE parallel to AB (I. 26.), bisect ED (I. 7.) in F, and draw FG parallel to AB, meeting the production of BC in G. B C G Because BC is equal to AE (I. 29.), BC and AD are together equal to AE and AD, or to twice AE and ED, or to twice AE and twice EF, that is,. to twice AF; consequently AF is half the sum of BC and AD. Wherefore the rectangle contained by the altitude of the trapezoid and half E F the sum of its parallel sides, is equivalent to the rhomboid BF (II. 1. cor.); but the rhomboid EG is equivalent to the triangle ECD (II. 7.), add to each the rhomboid BE, and the rhomboid BF is equivalent to the trapezoid ABCD. Cor. Hence the greater of two lines is equal to half their sum and half their difference; for AD is equal to AF joined to FD, which is half the difference ED. The smaller line AE again is formed by taking half the difference from half the sum. PROP. XIV. THEOR. The square described on the hypotenuse of a right-angled triangle, is equivalent to the squares of the two sides. Let ACB be a triangle which is right-angled at B; the square of the hypotenuse AC is equivalent to the two squares of AB and BC. For produce the base BA until AD be equal to the perpendicular BC, and on DB describe (I. 39.) the square DEFB, make GE and FH equal to AD or BC, join AG, GH, and HC, and through the points A and C (I. 26.) draw AL and CI parallel to BF and BD. E LH F Because the whole line BD is equal to DE, and a part of it AD equal to GE, the remainder AB is equal to DG; wherefore the triangles ACB and AGD are equal (I. 3.), since they have the sides AB, BC equal to DG, AD, and the contained angle ABC equal to ADG, both of these being right angles. In the same manner, it is proved, that the triangle ACB is equal to GEH, and to HFC. Consequently the sides AC, AG, GH, and HC are all equal. But G the angle CAB, being equal to AGD, is equal to the alternate angle GAL (I. 29.); add LAC to each, and the whole angle LAB or EDB (I. 29.) is equal to GAC, which is, therefore, a right angle. Hence the figure AGHC, having all its sides equal and one angle right, is a square. I D K B Again, the parallelograms KB and KE are evidently rectangular; they are also equal, being contained by equal sides; and each of them being double of the original tri angle ACB, they are together equal to the four triangles ACB, AGD, EHG, and HCF. The other inscribed figures LC and IA are obviously the squares of KC and AD, which are equal to the base and perpendicular of the triangle ABC. From the whole square DEFB, therefore, take away separately those four encompassing triangles, and the two interjacent rectangles KB and KE, and the remainders must be equal; that is the square AGHC is equal in space to both the squares ADIK and KLFC. Otherwise thus. Let the triangle ABC be right-angled at B; the square described on the hypotenuse AC is equivalent to BF and BI the squares of the sides AB and BC. For produce DA to K, and through B draw MBL parallel to DA (I. 26.) and meeting FG produced in L. Because the angle CAK, adjacent to CAD, is a right angle, it is equal to BAF; from each take away the angle BAK, and there remains the angle BAC equal to FAK. But the angle ABC is equal to AFK, both being right is equivalent to ABLK (II. 2. cor.), since they stand on equal bases AD and AK, and between the same parallels |