LESSONS IN OKOMLTIIY. 



i 



I.KSSONS IN GKoMi.i l:Y.- 



M XXVMI. To drat' tins late, the 



iig tides, and one of tk angles at the bate are 



lino A represent the length of the base of the 

 i t ri.ui/l.', n the Hum of it* remaining sides, nni the angle 

 tii" angles at its base. Draw any straight lino, x T, of 

 iiny point, i>, in it, make the angle 

 Y i) K equal to the given 

 B angle c. Then sot off 

 D r equal to A along D Y, 

 and D o equal to B 

 along D E, and join o r. 

 At the point r in the 

 straight line o F make 

 ! tho angle o r u equal 

 to the angle D o r, pro- 

 ducing F u, if neces- 

 sary, until it meeta the 

 eide D o of the triangle 

 T> o F in the point n. 



The triangle H F D ia the triangle required, for its base, D F, 

 ia equal u> the given straight line A, one of its angles F D H 

 is equal to the given angle c, and the length of its remaining 

 sides, D H, H F, taken together, is equal to B, for since the angle 

 H F o is equal to the angle u o F, H F is equal to u o, and 

 DO, or D H + H o, was made equal to B. 



Tito position of the point H in the straight lino D a may also 

 be found by bisecting F o in K, and drawing K L perpendicular 

 to F o, and cutting D o in H. 



PROBLEM XXIX. To draw a triangle having its angles equal 

 to tin' angles ofagiven triangle and its perimeter, or the sum of its 

 three sides, equal to a given straight line. 



Let the straight line A B represent the length of the perimeter, 

 or sum of the three sides of the required triangle, and CDS the 

 given triangle to whose angles the angles of the required triangle 

 must be equal. At the extremity A of the straight line A B make 

 the angle B A F equal to the angle E D c of the triangle c D E, 

 and at its extremity B, make the angle A B a equal to the angle 

 c E D. Bisect the angles B A F, A B G by the straight lines A H, 

 B K, and let these straight lines bo produced far enough to inter- 

 sect in the point L. From the point L draw L M parallel to A F, 

 meeting A B in u, and L N parallel to B o, meeting A B in N. 

 The triangle L M N thus formed is the triangle required, for it is 

 manifest that its angles at L, M, and N, are equal to the angles at 

 c, D, and F. of the triangle c D E, for the angle L M N, by Theorem 

 2 (page 156), is equal to the angle B A F, which was made equal 

 to the angle c D E, and the angle L N M, by the same Theorem, is 

 equal to the angle A B G, which was made equal to the angle c B D ; 

 and if there be two triangles each one of which has two angles 



which are equal to 

 two angles of the 

 other, the remaining 

 angle of the one 

 must be equal to the 

 remaining angle of 

 the other, since the 

 throe angles of 

 every triangle, whe- 

 ther great or small, 

 are together equal to 

 180 degrees; and as 

 in the triangle L H N 

 there are two angles L H N, L N M, equal to the angles c D B, 

 c E D of the triangle c D E, the remaining angle M L N of the 

 triangle L M N must be equal to the remaining angle D c E of 

 the triangle c D E. Now the side M L is equal to u A, be- 

 cause the angle M L A is equal to the angle M A L, M L A being 

 equal to L A F or H A F, because they are alternate angles, and 

 H A F being by the construction equal to u A H. For the 

 same reason the side N L of the triangle L M N is equal to N B. 

 Therefore the perimeter of the triangle i, :t v, or the sum of its 

 i, u, M N, N L, is equal to the given straight lino A B. 



XXX. To describe a square that shall be equal in 

 ,-ea to the sum of the squares described on two given 

 straight lines, 



Lot A and B be the two given straight lines ; it is required to 



Tig. 39. 



describe a square that shall be equal in superficial arm to the 

 squares described on these line*. First draw two -traigbt line* 

 of indefinite length, r Q, E s, intersecting each other at right 

 angles in the point c. On P and c Mt off c D, c , each 

 equal to A, and on c , c q set off c r, CO, each equal to B. 

 Complete the squares o D H K, c r K o, by Problem XVHL 

 (page 255) and join o . Upon a construct the square o E L M, 

 alo by Problem XVII I. The square o L K U equal in super- 

 ficial area to the squares c D a B, c r K o, described OB the giren 

 Htraight lines A and B respectively. 



Now at first sight it is difficult for any one who is ends*. 

 vouring by elf-tuition to acquire a knowledge of practical geo- 

 metry, whether for an agreeable change from other pursuits and 

 a useful mental exercise, 

 or to aid him in the pur- 

 suance of his calling 

 and there are many call- 

 ings, such as those of 

 the carpenter, mason, gar- 

 dener, wheelwright, etc., 

 in which a knowledge of 

 geometry is indispensable, 

 if ho who chooses any one 

 of them as the avocati >n 

 by which he must earn hts 

 daily broad wishes to rise 

 among his fellows, and 

 so deservedly command 

 the reward of his in- 

 dustry and intelligence 

 it may, we say, seem 

 at first difficult to 

 perceive that the largo 



Fig. 40. 



square G E L M is exactly equal in superficial area to the two 

 smaller squares c D H E, c F K G, taken together. We will, 

 therefore, first give him the means of proving to his satisfac- 

 tion, by the aid of his compasses and parallel ruler, that it is 

 so, and then endeavour, as in former cases, to deduce from a 

 consideration of Fig. 40 several geometrical facta that may be 

 gleaned from this problem, without the necessity of treating 

 them in separate problems. 



And first, for proof positive from ocular demonstration that 

 the area of the large square G B L H ia equal to the joint area of 

 the smaller squares c F K o, c D H B. An inspection of the 

 annexed figure, which is drawn on rather a smaller scale than 

 Fig. 40, but in precisely the same proportions, will ahow the 

 truth of the assertion. The two larger squares are divided into 

 their component parts in the following manner. Through c 

 draw c T parallel to G M or E L, meeting B o in T, in order to 

 fix the point T. Then through T draw T u parallel to c B, and 

 T v parallel to c o. Along T u set off T o equal to c o, and 

 through o draw o x parallel to T v or c o, meeting L M in x, and 

 through v draw v w, parallel to T u or c E, meeting o x in w. 

 Next, for the necessary division of the square c D H E, through 

 c draw c z, parallel to B G, and produce L B, to meet the straight 

 line D H in the point T. If this 

 figure be drawn on a piece of 

 paper, and the squares c F K o, 

 c D H B be cut out and divided, and 

 the pieces put together on the 

 square a E L M, so that the pieces 

 numbered 1, 2, 3, 4, 5, in the 

 smaller squares, be placed on the 

 divisions similarly numbered in 

 the Urge square, it will be found 

 that the area of the large square 

 is exactly equal to the joint area 

 of the smaller squares. 



It will bo noticed that the 

 straight lines p q, R 8 in Fig. 40 

 were drawn at right angles to each 

 other, and that the straight lines 

 c a, c E, that were set off along 

 c q, c B are at right angles to each other necessarily. Tin's is the 

 point in the construction on which the solution of the problem 

 depends, whatever may be the length of A and B, and to effect it 

 we have only to draw a line equal to A, and at right angles to 

 one end make a line equal to B, and join the extremities of the 



