GEOMETRY. 



fig. 9, where A B, CD, E F, and G H, 



being- all parallel to I K, are all parallels 

 to each other severally. 



N. B. They all make equal angles with 

 the oblique line O P. 



PROBLEM: x. 



To draw a parallel through a given point. 

 Fig. 10. From the end, on any part of the 

 given line AB, draw an oblique line to the 

 given point C. Measure the angle made 

 by ABC, and return another of equal 

 measurement upon the line B C, so as to 

 make the angle BCD equal to A B C : 

 the line C D will be parallel to the line 

 A B. Or, as in fig. 11, you may from any 

 points, say C D, in the line A B, draw two 

 semicircles of equal dimensions ; the tan- 

 gent E F will be parallel to A B. Or you 

 may, according to Problem 5, draw a per- 

 pendicular from the given point to the gi- 

 ven line, and draw another line through 

 the given point at right angles with the 

 perpendicular, proceeding from it to the 

 line whose parallel was to be made, and 

 which will be thus found. See fig. 12. 



THEOREM XI. 



Parallelograms of equal base and altitude 

 are reciprocally equal. Fig. 13. The pa- 

 rallelogram No. 1, is rectangular: No. 2 is 

 inclined, so as to hang over' a space equal 

 lo the length of its own base ; but the line 

 A. B, which is perpendicular thereto, di- 

 vides it into two equal parts ; let the left 

 half, A B E, be cut off, and it will, by be- 

 ing drawn up to the right, be found to fit 

 into the dotted space A C D. This theo- 

 rem might be exemplified in various 

 modes; but we presume the above will 

 suffice to prove its validity. 



THEOREM XII. 



Triangles of equal base and altitude are 

 reciprocally equal. Fig. 14. As every pa- 

 rallelogram is divisible into two equal and 

 similar triangles, it follows that the same 

 rule answers for both those figures under 

 the position assumed in this proposition : 

 we have shown this by fig. 15. 



PROBLEM X11T. 



To make a parallelogram equal to a given 

 triangle, with a given inclination or angle. 

 Fig. 16. Let B A C be the given triangle, 

 and E D F the given angle. On the line 

 D F measure a base equal to B C, the base 

 of the triangle. Take B G equal to half 

 the altitude of the triangle for the altitude 



of the parallelogram, and set it off on tfee 

 line E D. Draw F H parallel to E D, and 

 H E parallel to D F, which will complete 

 the parallelogram E F D H, equal to the 



PROBLEM XIV. 



To apply a parallelogram to a given right 

 line, equal to a given triangle, in a given 

 right line figure. Fig. 17. Let A B be the 

 given line to which the parallelogram is to 

 be annexed. Let C be the triangle tobe 

 commuted, and D the given angle. Make 

 B E F G equal to C, on the angle E B G ; 

 continue A B to E; carry on F E to K, 

 and make its parallel H A L, bounded by 

 F H, parallel to E A: draw the diagonal 

 H K, and G M both through the point B ; 

 then K L, and the parallelogram B M AL 

 will be equal to the triangle C, and be 

 situated as desired. 



PROBLEM XV. 



To make a parallelogram, on a given in- 

 clination, equal to a nght-linedfigure. Fig. 

 18. Let A B C D be the right-lined figure, 

 and F K H the given angle or inclination; 

 draw the line D B, and take its length for 

 the. altitude, F K, of the intended paralle- 

 logram, applying it to the intended base 

 line K M : now take half the greatest dia- 

 meter of the triangle D C B, and set it off 

 from K to M, and set off half the greatest 

 diameter of the triangle DAB, and set it 

 oft' from H to M : make G H to L M pa- 

 rallel to F K, and F G parallel to K H. 

 The parallelogram F K G H will be equal 

 in area to the figure A B C D, and stand at 

 the given inclination or angle. 



PROBLEM XVI. 



To describe a square on a given line. 

 Fig. 19. Raise a perpendicular at each 

 end of the line A B equal to its length ; 

 draw the line C D, and the, square is com- 

 pleted. 



THEOREM XVir. 



The square of the hypothenuse is equal to 

 both the squares made on the other sides of a 

 right -angled triangle. Fig. 20. This com- 

 prehends a number of the foregoing pro- 

 positions, at the same time giving a very 

 beautiful illustration of many. Let ABC 

 be the given right-angled triangle ; on 

 each side thereof make a square. For the 

 sake of arithmetical proof, we have assum- 

 ed three measurements for them : viz. the 

 bypothenuse at 5, onre other side at 4, attft 



