Chap. VIII.

GEOMETRY

879

To reduce Multilateral Figures to Triangles, as the regular pentagonA 13 C D E to a triangle. Prolong the base D C, and on it set oft’ twicethe length of DC to F and G ; from the centre H of the pentagon drawthe right lines II F and II G : the area of the triangle H G F will beequal to that of the pentagon.

To prove that the triangle H G F is equal to the pentagon A B C D E.

Draw from the centre II the two right lines II D and II C; then remarkthat the base F G of the triangle HGF contains five times the base ofthe triangle II C D, so that the triangle II G F is composed of five equaltriangles according to the 38th proposition of Euclid ’s First Book : butthe triangle II C D is also the fifth part of the regular pentagon A B C D E,so that the triangle HCD being the fifth part of the pentagonA B C D E, it is also of the triangle HGF: then, by the ninth propositionof Euclid’s Fifth Book, the triangle H G F is equal to the pentagonABODE.

Rectangles, it must be remembered, which have the same or equalaltitudes, are to one another as their bases: for if the base of one of therectangles be divided into any number of equal parts, the rectangle itselfwill be divided into as many equal rectangles by straight lines drawnparallel to its side through the points of division; also the base of theother rectangle will contain a certain number of parts equal to those intowhich the first base is divided, exactly or with an excess less than one of those parts, andthat rectangle will contain as many rectangles equal to those into which the first rectangle isdivided exactly, or with a corresponding excess less than one of them ; and this will be thecase, whatsoever be the number of parts in the first base and rectangle.

To reduce the Superficies of the irregular Pentagon IKLMN to a Triangle :

Fig. 11G8.

i?J

Fig. 11C9.

side of the pentagon on either hand asthe base ML: then from the op-posite angle to the side ML, as theangle I, draw the right lines IMand IL; from the point N draw N O,parallel to IM, and draw the line IO ;the triangle IPO will be equal tothe pentagon I K L M N. If the areaof the figure be reduced to an irre-gular hexagon, QRSTVX, youmust prolong one of the sides T S, on either hand, and from thepoint X draw the line X T, from the point V the right line V Y,parallel to XT, and from the point X draw the line XY, which,with the four other sides, will form an irregular pentagon,X Q R S Y, from which a triangle may be drawn according to thepreceding rule.

To prove that the triangle IPO is equal to the irregular pen-tagon IKLMN: it must be remarked that the two trianglesI XI N, I XIO, are equal, because they are on the same base I M,and between .the same parallels N O and I XI, so that if Hwe take the triangle I XI O for its equal IXIN, we shallhave the trapezoid IKLOequal tothepentagon IK L XI N ;the triangles IKL and I PL are also equal, so that inplacing the triangle I P L for its equal IKL, we have the Ftriangle IPO equal to the irregular pentagon I K L XI N.

To reduce an irregular Heptagon, as ABCDEFG, witha re-entering angle to a triangle : prolong the base E F oneach hand; draw from the point A a line A F, and fromthe point C another parallel to it, viz. G H, and draw H II :then from the angle C draw the right line C E, and fromD draw DI parallel to it, and then draw Cl; from thepoint B draw' B I, and from the point C the line CK, E

parallel to it, and draw B K ; from the point A draw A Kand B L parallel to it, and draw A L: the triangle A LIIwill be equal to the irregular heptagon ABCDEFG.

Any two rectangles are to one another in the ratio whichis compounded of the ratios of their sides : this is clearlyshown in the 23d Prop, of Euclid ’s Sixth Book ; and it alsois made apparent by the same proposition, that any twoparallelograms whatever are to one another in the ratiowhich is compounded of their bases and altitudes, for therectangles to which they are equal are in that ratio.

prolong any

tig 1171.