part of the given triangle, and is equiangular with it. Let D, E, F, be the bisections of the sides AB, BC, CA, respectively, of the given ▲ ABC; and let D, E, and E, F, and F, D, be joined: The ▲ DEF is one fourth part of the ▲ ABC, and is equiangular with it. For, join A, E; and, since (hyp.) BE=EC, CF=FA, and AD=DB, .. (E. 38. 1.) ▲ AEB AAEC; and the ▲ AEB is the double of the ▲ BDE, and the ▲ AEC is the double of the A CFE;.. the ABDE=A CFE; i.e. each of them is a fourth part of the A ABC; also they are upon equal bases BE and EC; .. (E. 40. 1.) DF is parallel to BC; and, in the same manner, it may be shewn that DE is parallel to AC, and FE parallel to AB; ., the figures FCED, DBEF, are ;.. (E. 34. 1.) the A DEFA DBE, which has been proved to be a fourth part of the G A ABC; also, the DFE, of the opposite opposite A DFE two B, and the FDE, of the BF,= DC, C; .. (E. 32. 1.) the DEF, of the the BAC, of the A ABC; and the ABC, DEF, are ... equiangular. 89. COR. 1. The straight line joining the bisections of any two sides of a ▲, is parallel to the remaining side. 90. COR. 2. If the four sides of any given quadrilateral rectilineal figure be bisected, the figure contained by the straight lines joining the several points of the bisection, shall be a parallelogram, which is the half of the given figure; also the four sides of this parallelogram shall be, together, equal to the two diagonals of the given figure. Let DH, HI, IF, FD be the straight lines joining the several bisections D, H, I, and F, of the sides AB, BG, GC, and CA, of the quadrilateral figure ABGC: The figure DHIF is a ; it is the half of the given figure ABGC; and its four sides are, together, equal to the two diagonals AG, BC, of the figure ABGC. First, since, D, H, F, I, are the bisections of the sides of the ABG, GCA, BAC, CGB, .. (S. 69. 1. cor.) DH and FI are parallel to AG, and DF and HI are parallel to BC; .. (E. 30. 1. DHIF is a: And, because DF is parallel to BC, and AB meets them, .. (E. 29. 1.) the ▲ ADL= 4 DBK; again, because DH is parallel to AG, and AB meets them, the DAL= 2 = BDK; and (hyp.) the side AD, of the A ADL, the side DB of the A DBK; .. (E. 26. 1.) DL=BK, LA= KD, and (E. 4. 1.) the ▲ ADL =▲ DBK; but DKEL being a □, DL = KE, and KD=EL (E. 34. 1.); .. BK= KE, and EL =LA: If, .., D, E be joined, the ▲ DLE = A DLA (E. 38. 1.) and the ▲ DKE= ▲ DKB; so that the KL the half of the A AEB, DK + FM AE, and DL + HN=BE. In the same manner it may be proved, that the LM the half of the AAEC, that the MN=the half of the A CEG, that the NK = the half of the A BEG, that LF+NI=EC, and that MI+KH= EG:.., the DHIF is the half of the given figure ABGC, and its four sides are, together, equal to the two diagonals AG, and BC. 91. COR. 3. It is manifest that the straight lines which join the opposite points of bisection of the sides of any trapezium, bisect each other. For, if D, I, and F, H, be the bisections of opposite sides of the given quadrilateral figure ABGC, it is manifest, from the preceding corollary, that the straight lines DI, FH which join them, will be the diameters of the DHIF; and.. (S. 42. 1.) they bisect one another. PROP. LXX. 92. PROBLEM. To describe a parallelogram, which shall be of a given altitude, and equiangular with, and also equal to, a given parallelogram. Let ABCD be a given, and E a given straight line: It is required to describe a which shall be equal to the ABCD, and also equiangular with it; and which shall have its altitude equal to the given line E. From the point C draw (E. 11. 1.) CF 1 to BC, and make CFE; through F draw (E. 31. 1.) HG parallel to BC; produce BA and CD to meet HG, in H and G; join H, C, and let HC cut AD in I; through I draw (E. 31. 1.) KIL parallel to HB or GC: The KLCG, which (constr. and E. 29. and 34. 1.) is equiangular with the ABCD, and has its altitude equal to E, is also equal to the ABCD. For, since BI and IG are compliments about the diameter HC of the HBCG, they are (E. 43. 1.) equal to one another; to each of these equals add the LD; and it is plain that the KLCG= ABCD. 93. COR. Hence, a rectangle may very readily be found, which shall be equal to a given square, and shall have one of its sides equal to a given straight line. PROP. LXXI. 94. THEOREM. If there be any number of rectilineal figures, of which the first is greater than the second, the second than the third, and so on, the first of them shall be equal to the last together with the aggregate of all the differences of the figures. First let there be three such given rectilineal figures. Make (E. 45. 1.) the FH equal to the greatest of the given figures, having its FGH of any given magnitude; produce GH to X; from HX cut off (E. 3. 1.) HIGH; find (S. 57. 1. cor.) a ▲ equal to the next greatest of the given figures, and apply (E. 44. 1.) to HI a equal to that A, having its IHK HGF: Again, from IX cut off IM=GH or HI, and, in like manner, to IM apply a IO, equal to the least of the given figures, and having its MIN=/ HGF. |