PROBLEM VIII. XII.] 



MATHEMATICS. PRACTICAL GEOMETRY. 



603 



angles to each Fig. 21. 



other ; in A B 



make A D 



equal to the N 



side of the 



square 1, and 



in A C make 



A E equal to 



the side of 



the square 2, 



and join DE; 



then in AB 



make A F 



equal to DE, 



and in A C 



make AG 



equal to the side of the square 3, and join F G ; again, 



in A B make A H equal to F G, and in A C, A K equal 



to the side of the square 4, and join H K ; lastly, in 



A B make A L equal to H K, and in A C make A M 



equal to the side of the square 5, and join L M ; then the 



square L M N O, described on the line L M, is equal to 



the sum of the squares 1, 2, 3, 4, 5. 



By proceeding in a similar manner till all the tides of any 

 number of given squares are employed, a square may 

 be constructed equal to the sum of the given number of 

 squares ; this, a well as the next problem, depending 

 throughout their whole construction on Euclid. Book I., 

 Prop. XLVIL 



PROBLEM VIII. 



To describe a square which shall be equal to the difference 

 of any two given squares (F and G). 



l>raw any two lines, A B and A C (Fig. 23), at right 



Fig. 23. 



angles to each other ; in 

 A C make A D equal to 

 the side of the less of the 

 two given squares, F ; 

 with the centre D, and a 

 radius equal to a side of 

 the other given square G, 

 describe an arc cutting 

 A B in E ; the square 

 A E H K, described on 

 the line AE, is the square 

 equal to the difference 

 of the two squares F 

 andQ. 



PROBLEM IX. 



To describe a square which shall be equal to a given 

 parallelogram (ABC D). 



Draw any line G K, which shall be a mean proportional 

 between the two sides, A R and B C of the given paral- 

 lelogram (Problem V.), and upon the line GK describe 

 Fig. M. 







the square HG K L, which will be equal to the given 

 parallelogram, A B C D. (Fig. 24). 



WTien the given parallelogram is not a rectangle, CM A E F B, 

 then draw a rectangle A B C D, standing on the same 

 base, and between the same parallels (Euclid, Book I., 

 Prop. XXXV.), eyiial to it, and proceed as before. 



PROBLEM X. 



To draw through a given point (E) a straight line, which 

 shall tend to the intersection of two given straight iitits 

 (A B, C D), but whose point of intersection (0) falls 

 beyond the limits of the drawing (L M N). 

 IST METHOD. Through the given point E (Fig. 20), 



draw any line E C, meeting A B in A, and C D in C, 

 draw any other line H K parallel to E C, meeting A B 



Fig. 25. 



L 



and C D in G and K ; join C G, and through K draw 

 K F parallel to it, meeting A B in F ; join. E G, and 

 through F draw F G parallel to it, meeting K G produced 

 in H ; join E H, which, if produced, would pass through 

 the point of intersection O. 



2ND METHOD. Draw any two parallels E C and G K. 

 (Fig. 26), as before ; join A K and C G, intersecting in 



F.f. 26. 



Fig. 27. 



F, and through F draw P Q parallel to E C or G K. 

 Join E K, cutting P Q in R, and through the point R 

 draw C H, meeting the line G K produced in H ; join 

 E H, which, if produced, would pass through the point 

 of intersection 0. 



PROBLEM XI. 



A t a given point (E) in a Bitten straight line (E D), to 

 make an angle equal to a given angle (A B C). (Euclid, 

 Book L, Prop. XXIII.) 



With the centre B (Fig. 

 27) and any convenient 

 radius, describe an arc, 

 cutting A B in H, and B C 

 in G ; with the centre E 

 and the same radius, de- 

 scribe an arc cutting E D 

 in the point L ; with the 

 centre L and a radius equal 

 to the distance H G, de- 

 scribe another arc, inter- 

 secting in the point K ; 

 join E K, and produce it if necessary ; then the angle 

 DBF will be equal to the angle A B C, as required. 



PROBLEM XH. 

 To bisect any given angle (A B C). (Euclid, Book I.. 



Prop. IX.) 



With the centre B (Fig. 28) and any convenient radius, 

 describe an arc cutting B A in D, and B C in E, with D 



Fig. 28. 



and E as centres, and any 

 equal radii, describe arcs in- 

 tersecting in the point F ; 

 join B F ; then the angle 

 ABC will be bisected by 

 B F, or divided into two 

 equal angles, A B F, and 

 CBF. 



By again bisecting the an- 

 gles A B F, CBF, the angle 

 ABC will be divided into 

 four equal angles, and so on ; by proceeding in a similar 

 manner, the angle ABC may be divided into any even 

 number of equal angles, continuing in geometrical pro- 

 gression viz., 2, 4, 8, 16, 32, 64, etc. 



