GEOMETRY. 



the last at 3. Now the square of 5 is 25. 

 The square of 4 is 16, and the square of 3 

 is 9 : it is evident the sum of the two last 

 sides make up the sum of the hypothe- 

 nuse's square ; for 9 added to 16 make 25. 

 But the mathematical solution is equally 

 simple and certain. The squares are let- 

 tered as follow : B D C E, F G B A, and 

 A H G K. Draw the following lines ; F C, 

 B K, A D, A L, and A E. We have al- 

 ready shown, that parallelograms and tin- 

 angles of equal base and altitude are re- 

 spectively equal. The two sides F B, 

 B C, are equal to the two sides A B, B D, 

 and the angle D A B is equal F B C : the 

 triangle A B D must therefore be equal to 

 the angle F B C. But the parallelogram 

 B L is double the triangle A B D, 'The 

 square G B is also double the triangle 

 F B C : consequently the parallelogram 

 B L is equal to the square G B, The 

 square H C in like manner is proved to be 

 equal to the parallelogram C L, which 

 completes the solution. Euclid, 47th of 

 1st Book. 



PROBLEM XVIIt. 



To divide a line, so that the rectangle con- 

 tained under the -whole line and one segment 

 be equal to the square of the other segment. 

 Fig. 21. On the given line A B describe 

 the square A B C D ; bisect A C in E, and 

 with the distance E B extend A C to F, 

 measuring from E. Make on the excess 

 F A the square F H, and continue G H to 

 K. The square F H will be equal to the 

 parallelogram H D. 



PROBLEM XIX. 



To make a square equal to a given riglti- 

 linedjigure. Fig. 22- Let A be the given 

 right-lined figure : commute it to a paj-ql- 

 lelogram, B D, as already shown, (prob. 

 15.) : add the lesser side E D to B E, so 

 as to proceed to F : bisect T> F in G, and 

 from that point describe the semicircle 

 B II F. Continue D E to H, which will 

 give If E for the side of a square equal in 

 area to the parallelogram B D, and to the 

 original given figure A. 



PROBLEM XX. 



To find the centre of a given circle. Fig. 

 23. Draw at pleasure the chord A B, bisect 

 it in D by means of :t diameter, which 

 being bisected will give F for the centre 

 of the circle. 



PROBLEM XXI. 



To complete a circle upon a given segment- 



Fig. 24. Let A B C be the'given segment: 

 draw the line A C, and bisect it in D ; 

 draw also the perpendicular B E through 

 D, draw B A, and on it make the angle 

 B A E, equal to D B A ; this will give the 

 point of intersection E for the centre, 

 whence the circle may be completed. It 

 matters not whether the segment be more 

 or less than a semicircle. 



PROBLEM xxir. 



To cut a given circumference into tivo 

 equal parts Fig. 25. Draw the line A B, 

 bise-ct in C ; the perpendicular D C will 

 divide the figure into two equal and simi- 

 lar parts. 



PROBLEM XXI H. 



In a given circle to describe a triangle 

 eqaiangidar to a given triangle. Fig. 26. 

 Let A B C be the circle, andD E F the tri- 

 angle given. Draw the line G H, touch- 

 ing the circle in A: make the angle HAG 

 equal to D E F, and GAB equal to D F E; 

 draw B C, and the triangle B A C will be 

 similar to the triangle D E F. 



PROBLEM XXIV. 



Jibout a given circle to describe a triangle 

 similar to a given triangle. Fig. 27. Let 

 A B C be the given circle, and D E F the 

 given triangle : continue the line E F both 

 ways to G and H, and having found the 

 centre K, of the circle, draw a radius, 

 K B, at pleasure ; then from K make the 

 angle B K A equal to D E C, and B K C' 

 equal to D F H ; the tangents L N per- 

 pendicular to K C, M N perpendicular to 

 K B, and M L perpendicular to K A, will 

 form the required triangle. 



PROBLEM XXV. 



To describe a circle about a given trian~ 

 gle. Fig. 28. In the given triangle ABC, 

 bisect any two of the angles ; the inter- 

 section of their dividing lines, B D and 

 C D, will give the centre D, whence a 

 circle may be described about the trian- 

 gle, with the radius D C. 



PROBLEM XXVI. 



To Inscribe a circle in a given triangle. 

 Fig. 29. In the triangle ABC, divide the 

 angles A B C, and B C A, equally by the 

 lines B D, C D. Their junction at D will 

 give a point whence the circle E C F may 

 be described, with the radius D F per- 

 pendicular to B C. 



