Chap. VIII.

GEOMETRY.

887

2’o divide Pentagons by lines parallel to their sidesinto several equal parts, as the irregular pentagon,

ABC1) E, into two equal parts, by a line parallelto E D: reduce the pentagon to a triangle, A F G;bisect its base G F in H, and draw A H : reduce thetrapezoid, AI1DE, to a triangle, EID; prolong thebase I D, and the side A E, until they intersect in K:find a mean proportional, L M, between K I and KD;set off the mean proportional on K I, from K to N;through the point N draw a line, N O, parallel toE D, and it will divide the irregular pentagon into twoequal parts.

7b divide Hexagons into several equal parts, uniting at one of their angles : first reducethe irregular hexagon, as A B C D E F, to a pentagon: then to a trapezoid, and lastly toa triangle. To reduce the hexagon, A B C D E F, to apentagon, prolong the side D C to I: draw A C, and fromB a line parallel thereto B K, and divide A K, which willreduce the hexagon to the pentagon, A K D E F. Toreduce this to a trapezoid, prolong the side E D, and drawA D, and from K draw K G parallel to A D ; tlraw A G,which will reduce the pentagon to the trapezoid, A G E F:reduce the trapezoid to a triangle by drawing from A theright line A E, and through F its parallel F H, and drawA II, which will form the triangle A G II, equal to the trapezoid A G E F: then dividethe base, HG, into two equal parts in L, and draw A L, which will divide the hexagon intotwo equal parts.

To prove that the irregular hexagon ABCDEF is divided into two equal parts ABCDLand ALEF, which join at the point A.

The angle AG F is equal to the irregular hexagon ABCDEF, and the triangle A G Hhas its base equally divided at the point L, so that the two triangles AGL and ALH,having the same height and base, are equal; and as they are the moieties of the triangleA GH, they are also the moieties of the hexagon A B C D E F, which is equal to thetriangle A G II.

The two triangles AE F and A E II being upon the same base A E, and between thesame parallels FH and A E, they are equal: AEF and AEH are also equal, as areFAZ and HEZ.

A

Fig. 1207.

on H N C F

Fig. 1206.

If from the triangle ALH, which is the half of the hexagon ABCDEF, we cut offthe triangle HEZ, and take its equal FAZ, we shall have the figure ALEF equal to thetriangle A L H, and also the moiety of the figure A B C D E F.

7b divide Multilateral Figures, having re-entering angles, into several equal parts, unitingat one angle, as to divide the irregular heptagon, ABCDEFG, into six equal parts,uniting at the angle B : reduce theirregular heptagon to a triangle B N M,draw B D in C N; divide M N intosix equal parts, in the points 1, 2, 3, 4, 5, 6,and draw right lines to them from B ;draw E B, and through 2 a line 2 O,parallel to E B, and draw O B. Reducethe irregular pentagon A B O F G to atrapezoid, B O F P, by prolonging F G,and from G draw the right line G B,and from A, A P, parallel to B G, anddraw P B : set off the base of one of thesmall triangles, as 4 3, and F' G, fromP to Q. Take also the length of itsside 4 B, and describe from the pointP an arc R ; take the length of its otherside, 3 B, and from the point A describethe second arc S, cutting the first in T: m i 2 k 3 4 0 d n «

then draw the two lines Q T and P T, p ig 12 08which will form the triangle Q P T

equal to B 4 3. Through B draw a line parallel to F Q, as B V, cutting P T in X : thenlower the triangle T P Q to X, and you will have X P Y equal to T P Q. If, then, youset off the base Y P from P to Z, and draw Z B, you will have the two triangles B Z Pand X P Y equal to each other. The triangles BG A and 15 G P being equal, if to thetriangle BZ G you add the triangle B G P , you will have the triangle BZP; and if to thetriangle B Z G you add B G A, you will have the trapezoid B Z G A equal to B Z P, and

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