(17.) If from any angle of a rectangular parallelogram a line be drawn to the opposite side, and from the adjacent angle of the trapezium thus formed another be drawn perpendicular to the former; the rectangle contained by these two lines, is equal to the given parallelogram. B A From A, one of the angles of the rectangular parallelogram ABCD, let any line AE be drawn to the opposite side, cutting off the trapezium ABCE; from B let fall the perpendicular BF; the rectangle contained by AE, BF, is equal to the parallelogram ABCD. For BA being parallel to ED, the angle BAF = AED, and the angles at F and D are right angles, .. the triangles BAF, AED are similar; whence BF : BA :: AD : AE, and the rectangle AE, BF is equal to the rectangle BA, AD, i. e. to ABCD. (18.) To divide a parallelogram into two parts which shall have a given ratio, by a line drawn parallel to a given line. B L Let ABCD be the given parallelogram, and EF the line whose direction is given. Divide AD in G in the given ratio, and make GH=AG. Join BH; bisect it in I, and through I draw KL parallel to EF; KL divides the parallelogram in the given ratio. For (Eucl. vi. 1. and i. 41.) AKG H E ABH ABCD :: AG: AD. But the triangles BIL, KIH are equal, since BI=IH, and BL is parallel to KH; .. ABH = ABLK, and ABLK: ABCD :: AG: AD, :. ABLK : LKDC :: AG : GD, i. e. in the given ratio. (19.) To bisect a trapezium by a line drawn from one of its angles. Let ABCD be the given trapezium, and Α the angle from which it is to be bisected. Draw the diagonals AC, BD; and bisect BD, which is opposite to the angle A, in the point E. Join AE, CE; and through E draw FEG parallel to AC. Join AG; AG bisects the trapezium. Since DE is equal to EB, the triangles AED, AEB are equal; as also DEC, BEC; .. the figure AECD is equal to the figure AECB. Also (Eucl. i. 38.) the triangles AEG, CEG are equal; take away ... the common part EHG, and AEH= GHC. To the figure AECD add AEH, and take away its equal GHC; and to AECB add GHC, and take away AEH; and the triangle AGD is equal to the trapezium AGCB, or the given trapezium is bisected by AG. (20.) To bisect a trapezium by a line drawn from a given point in one of its sides. Let ABCD be the given trapezium, and P the given point. Join PA, and from the angle P bisect the trapezium APCD (iv. 19.) by the line PE. On PE make the triangle PEF equal to ABP. Bisect EF in G; join PG. PG bisects the trapezium. DFGE Since FG is equal to GE, the triangle PGF is equal to the triangle PGE. But PGE is equal to half the triangle ABP; and PEC is half the figure PABC; whence PGC is half of the trapezium ABCD; which is .. bisected by PG. (21.) 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 half the trapezium. Let ABCD be a trapezium, having the side AB parallel to DC. Let AD be bisected in E; join BE, CE; the triangle BEC is half of the trape zium. B E G Through E draw FEG parallel to BC, meeting CD in G, and BA produced in F. The alternate angles FAE, EDG being equal, as also the angles at E, and AE=ED, . the triangles AEF, DEG are equal; whence the parallelogram BFGC is equal to the trapezium ABCD. But BFGC and the triangle BEC, being on the same base BC, and between the same parallels BC, FG, the triangle BEC is half of BFGC, and .. also half of ABCD. COR. From the demonstration it appears, that a trapezium which has two sides parallel, may be reduced to a parallelogram equal to it, by drawing through the point of bisection of one of the sides, which are not parallel, a line parallel to the other of those sides, and meeting the parallel sides. (22.) To divide a given trapezium whose opposite sides are parallel, in a given ratio, by a line drawn through a given point, and terminated by the two pa G in the given ratio; and through G and P draw IPH; IPH will divide the trapezium in the given ratio. Draw KGL, MFN parallel to AD; then EA=KG =MF, and ED=GL=FN; but AE=ED, :. KG= GL, and MF = FN; whence (iv. 21. Cor.) ADIH = AL, and HICB=KN. Now AL KN :: EG: GF, .. ADIH : HICB:: EG: GF, i. e. in the given ratio. (23.) If a trapezium, which has two of its adjacent angles right angles, be bisected by a line drawn from the middle of one of those sides which are not parallel; the sum of the parallel sides will have to one of them the same ratio, that the side which is not bisected has to that segment of it which is adjacent to the other. Let ABCD be a trapezium, having the angles at A and D right angles, and .. AB, DC parallel; and let the trapezium be bisected by EF; if AD be bisected in E, AB+DC: AB :: BC: CF; but if BC be bisected in F, AB+DC: AB :: AD : DE. E H B I Produce DA, CB to meet in G. Join AF, DF, BE, CE, and let fall the perpendiculars AH, DI. Since AE=ED, the triangles AFE, DFE are equal, .. the triangles DFC, AFB are equal; .. the rectangles FC, DI and BF, AH are equal, and FC FB :: AH : DI :: AB : CD, ·. AB+CD : AB :: (FC+FB=) BC : FC. EBF, ECF being equal, the triangles .. (Eucl. vi. 15.) AB: CD :: DE: EA, E D B and AB+CD: AB :: (DE+EA=) AD: DE. In like manner AB+ DC : DC :: AD : AE. (24.) If the sides of an equilateral and equiangular pentagon be produced to meet; the angles formed by these lines are together equal to two right angles. Let ABCDE be an equilateral and equiangular |