880

THEORY AND PRACTICE OF ENGINEERING.

Book II.

7b reduce a Square to a Parallelogram on a given length,as the square A B C D to a parallelogram which shall beof the length E F : prolong the side A of the square, andon it set the given length E F, from B to G, and prolongthe other three sides; draw the right line GC till it cutsAD prolonged in H: through the point II draw a lineparallel to A G, as H I, and from G draw a line parallel tothe side B C I, as is GL, cutting the side C I) prolongedin M : then the parallelogram CMLI will be equal to thesquare A B C D.

Fig. 1172.

The complements of the parallelograms which are about thediameter of any parallelogram are equal to one another: letII A GL be a parallelogram, of which the diameter is II G ;and D I, MB parallelograms about HG, that is through whichIIG passes; and A C , C L, the other parallelograms, whichmake up the whole figure H A G L, which are therefore calledthe complements. The complement A C is equal to the com-plement CL, because IIA GL is a parallelogram, and II Gits diameter ; the triangle H A G is equal to the triangle H L G ;and because D C III is a parallelogram, the diameter of which isIIC, the triangle II DC is equal to the triangle HIC; by thesame reason the triangle CB G is equal to the triangle CMG:then because the triangle II1) C is equal to the triangle III C,and the triangle CBL to CMG, the triangle IIDC, togetherwith the triangle CB G, is equal to the triangle III C, togetherwith the triangle CMG: but the whole triangle II AG is equalto the whole II L G ; therefore the remaining complement A C isequal to the remaining complement CL. Fig. 1173.

To reduce a Rectangle to a Square _Prolong one of the long sides of the rectangle, as

A B to II, and set off the width A G on it, from A to E : to find a mean proportionalbetween A and A E, divide E B in F, and from the point F, with the radius FII, describea semicircle: prolong AD till it cuts the semicircle in G : AG will be the mean pro-portional, and the square A C I H constructed on it will equal A B C D.

To lengthen or shorten a Parallelogram on a given

length. It is required to reduce the parallelogram A B C D, F _ F

to one which shall be of the length E F: prolong thefour sides of the parallelogram to infinity, and set off thegiven length from B to G, and from C to II, and thendraw GII parallel to B C : draw the diagonal G C, tillit cuts A D prolonged in I, and draw through this pointII, a line parallel to D C H, cutting B C prolonged in K,and G H in L *, then the parallelogram CII L K. will be 11 1 L

equal to A B C D. p ig> 1174

The straight lines which join the extremities of twoequal and parallel straight lines towards the same parts are also themselves equal andparallel: let BG and CM be equal and parallel straight lines, and joined towards the sameparts by the straight lines BC, GM; BC, G M arc also equal anrl parallel; join GC: andbecause BG is parallel to C M, and C G meets them, the alternate angles BGC, GC M areequal; and because BG is equal to CM and GC common to the two triangles BGC, MCG,the two sides BG, GC are equal to the two M C, C G, and the angle B G C is equal to theangle G C M : therefore the base BC is equal to the baseGM, and the triangle BGC to the triangle GCM, andthe other angles to the other angles, each to each, towhich the equal sides are opposite : therefore the angleB C G is equal to the angle C G M, and because thestraight line GC meets the two straight lines B C, G M,and makes the alternate angles BCG , CGM equal to oneanother, B C is parallel to GM, and it was shown to beequal to it.

To reduce the Area of a Circle to that of a Square , §*c. Itis required to reduce the circle H I K to a square: dividethe diameter K I of the circle into 14 equal parts, andcount off 11 of these parts to L ; at this point L, and inK I, elevate a perpendicular LM; remark where it cutsthe circle in H, and draw K H ; then a square con-structed as KII, as is NO IIK, will be equal to thecircle H I K.

Fig.1175.