and DF and having the same altitude, contain equal spaces. For let the bases AC, DF be placed in the same straight line, join BH, and produce it both ways, draw AG and DH parallel to CB and FE (I. 26.), and join AH, CE. Because the triangles ABC, DEF are of equal altitude, GE is parallel to AF (I. 27.), and B H GC, HF are parallelograms. But C AC, being equal to DF, and DF, equal (I. 29.) to HE, must also be equal to HE, and, therefore, (I. 30.) AE is a rhomboid or parallelo gram. Whence the rhomboid GC is equivalent to AE (II. 1.), and this again is, for the same reason, equivalent to HF; consequently GC is equal to HF, and therefore their halves or (Cor. I. 25.) the triangles ABC and DEF are equivalent. Cor. Hence rhomboids on equal bases and between the same parallels, are equivalent. 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 joining their vertices, is parallel to AC. B For if BD be not parallel to AC, draw the parallel BE meeting a side of the triangle ADC in E, and join CE. Because BE is made parallel to AC, the triangle ABC is equivalent to AEC; (II. 2.) 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. 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. THEOR. A straight line bisecting two sides of a triangle, is parallel to the base. The straight line DE, which joins the middle points of the sides AB, BC, is parallel to the base AC of the triangle ABC. B For join AE and CD. Because the triangles ACD, BCD stand on equal bases AD, DB, and have the same vertex or altitude, they are (II. 2.) equivalent, and, therefore, ACD is half of the whole triangle ABC. For the same reason, since CE is equal to EB, the triangle CAE is equivalent to EAB, and is consequently half of the whole triangle ABC, Whence the triangles ADC and AEC are equal; and they stand on the same base, and have, therefore, the same altitude (II. 3.), or DE is parallel to AC. A D E Cor. Hence the triangle DBE cut off by the line DE is the fourth part of the original triangle. For bisect AC in G, and join DG, which is, therefore, parallel to BC. The triangle ADG is equivalent to GDC (II. 2.), and GDC, being the half of the rhomboid GE, is equivalent to DEC, which again is (II. 2.) equivalent to DEB. The triangle ABC is thus divided into four equivalent triangles, of which DBE is one. Hence also the rhomboid GDEC is half ofthe original triangle. PROP. V. THEOR. Straight lines joining the successive middle points of the sides of a quadrilateral figure, form a rhomboid. If the sides of the quadrilateral figure ABCD be bisected and the points of section joined in their order; EFGH is a rhomboid. For draw AC, BD. And because FG bisects AB, BC, it is, by the last Proposition, parallel to AC; and for the same reason, EH, as it bisects AD and DC, is parallel to AC. Wherefore FG is parallel to EH (I. 32.). In like man B C Κ. H E ner, it is proved that EF is parallel to HG; and consequently the figure EFGH is a rhomboid or parallelogram. Cor. Hence the inscribed rhomboid is half of the quadrilateral figure. For IG is half of the triangle ABC (II. 4. cor.), and IH is half of the triangle ADC. Consequently the rhomboid EG is half of the whole quadrilateral figure ABCD. PROP. VI. PROB. To find a triangle equivalent to any rectilineal figure. Let it be required to reduce the five-sided figure ABCD 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, and 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, and 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 C 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 away from the space ACDE, and there remains the quadrilateral figure FCDE equal 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 equal to the quadrilateral figure FCDE, and is consequently equal 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. PROP. VII. 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. For join BD and EC. The triangles ABD and DBC having the same vertex and equal bases, are (II. 2.) equivalent. But the B diagonal E bisects the rhomboid DEFC F (Cor. I. 29), and the triangles DBC A and DEC, having the same altitude, are equivalent (II. 1.); consequently their doubles, or the triangle ABCD and the rhomboid DEFC are equivalent. PROP. VIII. 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. B E F Reduce the rectilineal figure to an equivalent triangle ABC (II. 6.), bisect the base AC in the point D (I. 7.), and draw DE making an angle CDE equal to the given angle G (1. 4.), through B draw BF parallel to AC (I. 26.), 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 an angle CDE equal to G, and is equivalent to the triangle ABC (II. 7.), and consequently to the given rectilineal figure. PROP. IX. THEOR. Equiangular rhomboids contained by equal sides are equal. |