Let E be any point in the ABCD, and let E, A, and E, B, and E, C and E, D be joined: The two AEB, DEC are, together, half of the ABCD. For, through E draw (E: 31. 1.) FEG parallel to AB or DC: and since AG, and GD are, .. (E. 41. 1.) the ▲ AEB is the half of the and the ▲ DEC is the half of the ▲ AEB+A DEC is the half of the AG, GD; .. the AG + the half of the GD, or the half of the whole ABCD. PROP. LXVIII. 87. THEOREM. If two sides of a trapezium be parallel, the triangle contained by either of the other sides, and the two straight lines drawn from its extremities to the bisection of the opposite side, is the half of the trapezium. Let the two sides FD, EC, of the trapezium FECD be parallel; let K be the bisection of either of the two remaining sides, as DC; and let K, E and K, F be joined:. The A FKE is the half of the trapezium FECD. = For, through K draw (E. 31. 1.) LKM parallel to FE, and let LKM meet BC in M and FD, produced, in L. And since FL (constr.) is parallel to EC, and LM meets them, .. the DLK= z KMC; also (E. 15. 1.) the ▲ DKL=2CKM, and (hyp.) the side DK of the ▲ DKL, the side CK of the A CKM ; ... (E. 26. 1. and E. 4. 1.) the ▲ DKL 'A CKM; but if to the rectilineal figure FEMKD there be added the A CKM, there results the trapezium FECD; and if to the same figure there be added the ▲ DKL, there results the FEML; .. these results are equal; but (E. 41. 1.) the A FKE is the half of the FEML; .., the ▲ FKE is the half, also, of the trapezium FECD. 88. THEOREM. PROP. LXIX. The triangle contained by the straight lines joining the points of the bisection of the three sides of a given triangle, is one-fourth 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 A 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.) A AEB AAEC; and the ▲ AEB is the double of the A BDE, and the A AEC is the double of the A CFE; .. the ABDE= A CFE; i.e. each of them is a fourth part of the ▲ 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 DEF = A DBE, which has been proved to be a fourth part of the G A ABC; opposite opposite ▲ DFE two also, the DFE, of the C; .. (E. 32. 1.) the BF,= DC, = DEF, of the the BAC, of the ▲ 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 2 DBK; again, because DH is parallel to AG, and AB meets them, the DAL = BDK; and (hyp.) the side AD, of the ▲ ADL, the side DB of the ▲ DBK; .. (E. 26. 1.) DL=BK, LA= KD, and (E. 4. 1.) the ▲ ADL =A 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 ▲ 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 BÉG, 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 |