PROP. III. THEOR. Equivalent triangles on the same or equal bases, have the same altitude. If the triangles ABC and ADC, standing on the same base AC, contain equal spaces, they have the same altitude, or the straight line which joins their vertices is parallel to AC. B. For if BD be not parallel to AC, draw the parallel BE meeting AD or that side produced, in E, and join CE. Because BE is made parallel to AC, the triangle ABC is (II. 1.) equivalent to AEC; but ABC is by hypothesis equivalent to ADC, and therefore AEC is equivalent to ADC, which is absurd. The supposition then that BD is not parallel to AC involves a contradiction. E The same mode of demonstration, it is obvious, will apply in the case where the equivalent triangles stand on equal bases. Cor. Hence equivalent rhomboids on the same or equal bases, have the same altitude. PROP. IV. PROB. To find a triangle equivalent to any rectilineal figure. Let it be required to reduce the five-sided figure ABCDE to a triangle, or to find a triangle that shall contain an equal space. Join any two alternate points A, C, and through the intermediate point B, draw BF parallel to AC, meeting either of the adjoining sides AE or CD in F; which point, when the angle ABC is re-entrant, will lie within the figure: Join CF. Again, join the alternate points C, E, and through the intermediate point D draw the parallel DG, to meet in G either of the adjoining sides AE or BC, which, since the angle CDE is salient, must for that effect be produced; and join CG. The triangle FCG is equivalent to the five-sided figure ABCDE. A F E D Because the triangles CFA and CBA have by construction the same altitude and stand on the same base AC, they are (II. 1.) equivalent; take each of them away from the space ACDE, and there remains the quadrilateral figure FCDE equivalent to the five-sided figure ABCDE. Again, because the triangles CDE and CGE are equal, having the same altitude and the same base; add the triangle FCE to each, and the triangle FCG is equivalent to the quadrilateral figure FCDE, and is consequently equivalent to the original figure ABCDE. In this manner, any polygon may, by successive steps, be reduced to a triangle; for an exterior triangle is always exchanged for another equivalent one, which, attaching itself to either of the adjoining sides, coalesces with the rest of the figure. Schol. This problem is of singular use in practice, since it enables the surveyor greatly to abridge his computations, by reducing any plan that he has delineated at once to an equivalent triangle. PROP. V. PROB. A triangle is equivalent to a rhomboid which has the same altitude and stands on half the base. The triangle ABC is equivalent to the rhomboid DEFC, which stands on half the base DC, but has the same altitude. B For join BD and EC. The triangles ABD and DBC having the same vertex and equal bases, are (II. 2. cor. 2.) equivalent. But the diagonal EC bisects the rhomboid DEFC (I. 26. cor.), and the triangles DBC and DEC, having A the same altitude, are equivalent (II. 1.); consequently their doubles, or the triangle ABC and the rhomboid DEFC, are equivalent. Cor. Hence the area of a triangle is equal to half the rectangle contained under its base and its altitude-from which property is derived the mensuration of any rectilineal figure. PROP. VI. PROB. To construct a rhomboid equivalent to a given rectilineal figure, and having its angle equal to a given angle. Let it be required to construct a rhomboid which shall be equivalent to a given rectilineal figure, and contain an angle equal to G. Reduce the rectilineal figure to an equivalent triangle ABC (II. 4.), bisect the base AC in the point D (I. 7.), and draw DE making an angle CDE equal to the given angle G (I. 4.), through B draw BF parallel to AC (I. 23.), and through C the straight line CF parallel to DE: DEFC is the rhomboid that was required. For the figure DF is by construction a rhomboid, contains E F an angle CDE equal to G, and is equivalent to the triangle ABC (II. 5.), and consequently to the given rectilineal figure. PROP. VII. 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. Since the diagonal AF bisects the rhomboid EI (I. 26. cor.), the triangle AEF is equivalent to AIF; and for the D gonal, 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 AIF and FGC, and there remains the rhomboid BF equivalent to FD. Cor. The same property will extend to the spaces left on both sides of the diagonal by rhomboids any how combined. PROP. VIII. PROB. With a given straight line to construct a rhomboid equivalent to a given rectilineal figure, and having an 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. E B Κ Construct (II. 6.) the rhomboid BF equivalent to the given 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 extension of BH in C, join CF and produce it to meet the extension of BE in A; draw AD parallel to EF, meeting CG in D, and produce HF to I: FD is the rhomboid required. A For FD and FB are evidently complementary rhomboids, and therefore (II. 7.) equivalent; and, by reason of the parallels AE, IF, the angle FID is equal to EAI (I. 22.), which again is equal to BEF or the given angle K. Schol. This problem might also be solved by repeated operations; each triangle, into which the rectilineal figure is divided, being successively converted into a rhomboid, having an angle equal to K, and placed on a line equal to L, or the summit of each preceding rhomboid. These rhomboids will evidently coalesce and fulfil the conditions required. The process is not so direct as when the figure was previously reduced to an equivalent triangle; but it seems better adapted for the solution of another similar |