Chap. VIII.

GEOMETR V.

885

To divide the Triangle I'-TG into four equal partsmeeting at the point H, on the side FG: divide GFinto four equal parts in the ppints 1,2,3,4 ; from the pointK draw the right line E 1 ; from the point H draw theright line II1, and through E a line parallel to it,cutting G F in K, and draw II K : set off the line G Kon G I’ from K to L, and draw H L: set off K L fromL to M on G F, and draw H M and H F ; draw M Nparallel to it, cutting EF in () ; draw HO; the linesH O, II L, and H K, will divide the triangle E FG intofour equal parts.

To prove that the triangle E F G is divided into fourequal parts, II K G, H L K, H O F L, II E O, it is only

Fig. 1196.

\

requisite to remark that the triangle E F G has its base G F divided into four equal partsby the points 1, 2, 3, and 4, and which form the triangles E I G, E 2 I, E 3 2, and E F 3,the heights and bases of which are all equal, and each forms a fourth of the triangle E F G.This must be so, for the triangles II E K and I E K are upon the same base E K, andbetween the same parallels II l and E K, so that cutting off the common triangle P E K,the two triangles IIEP and I K P, winch are equal, will remain; if from the triangleE I G we cut off the triangle IIEP to take its equal I K P, we shall have the triangleI1KG equal to E I G, and consequently to a quarter of the triangle E F G.

It is to be observed that the three triangles II KG, II LK, and I1ML, being of thesame height, and having their bases G K, K L, and L M equal, are equal each to each ; andas the triangle II K G has been proved to be the fourth of the great triangle E F G, the twoother triangles H LK and II ML are also each the fourth of this great triangle: but asthe triangle H M L comes out of the triangle E F G, that it may be comprised within it,the right line II F, and its parallel M N are drawn, which form the triangles H O M andF () M, which being on the same base () M, and between the same parallels H F and O M,are equal, so that in cutting off from these two equal triangles the common triangle Q O M there will remain the two triangles IIOQ and FMQ, which are equal; if from the

triangle H M L wc cut off the triangle F M Q, to take its equal H () Q, we have the figure

II O FL equal to the triangle H ML, and dividing into four the great triangle EFG; andas we have proved that the three figures IIKG, |

II LK, and II O F L, are each a quarter, J

the remainder H E O is the fourth quarter.

To divide the Triangle A13 C into threeequal parts uniting at the point D: drawthrough the point A a line AF parallel tothe line BC; divide C13 into three equalparts in the points 1, 2, 3; draw AD, andthrough 1 a line 1 F, parallel to A D, cuttingA 13 in G : from I) draw 1) G and D 1 : set y c~ 3 ~

off 131 from B to II, and D G from II to I,and on this line III draw the triangle K III g *

equal and similar to A G 1) : elevate the triangle Killto the height of ABC, and you will have LM II : setofl* II M from B to N, and H N from A to C, and drawO C: take the shortest distance from D to A C, as1) P, to draw a line parallel to A C at the distance, asQ R, cutting A E in S ; from which point S, draw S C,and through C another parallel to it, as CT, cuttingAC in V' ; draw V D : the lines 1 D, D G, and I) Vwill divide the triangle ABC into three equal partsuniting at the point I).

To divide Triangles into equal Parts by lines parallel totheir sides: as to divide A 13 C into two equal partsparallel to A C; prolong one side C B, and bisect it inD ; set off B D from B to E: find a mean proportionalbetween CBand BE, as BH : set off BH on J3C fromB to I: through I draw I K parallel to AC: the lineI K will divide the triangle ABC into two equal parts.

To divide Figures of four Sides into several equal parts,as the square A 13 C D into three equal parts, all unitingat A : draw the diagonal 13 1), and divide it into threeequal parts, in 1,2,3: from A draw the line Al, A2:draw A C and 1 E and 2 F parallel to A C : then drawA E and A F; these two lines will divide the squareA BCD into three equal parts uniting in the angle A.

3 l 3

n N H M I

Fig. 1198.

Fig. 1199.