. I II. From the given point (1) draw a perpendicular to the given line (2, 3). (It will be found that whatever point be taken on the side remote from (1), the circle drawn will not cut (2, 3): so produce (2, 3) both ways to cut the circle.) III. From the point O draw a perpendicular to the line PQ. Particular Enunciation. Given. The line A B. The point A in it. The rectilineal angle DCE. 2 Required. At the point A in the line AB, to make a rectilineal angle equal to the given angle DCE. 3 PROBLEM (Euclid I. 23). Repeat the enunciation of Euc. I. 8 (page 46). General Enunciation. C At a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle. P Construction. In CD, CE take any C points D and E. Join DE. By means of Euc. I. 22 construct the triangle AFG, (a) having its side AF equal to CD; its side A G equal to CE; and its side FG equal to DE. Then the angle FAG shall be equal to the angle DCE. B Proof (that angle FA G equals angle D CE.) If two triangles have two sides, and the base of one, equal to two sides and the base of the other, each to each. The angles included between those sides are equal. CDE and AFG are two triangles having the two sides DC, CE equal to the two sides FA, AG, each to each, and the base DE equal to the base FG (by Construction a). .. The included angle F A G is equal to the included angle D C E. Result.-Wherefore at the given point A in the straight line AB, the angle FA G has been made equal to the given rectilineal angle D C E. Q. E. F. EXERCISES.-I. Write out this proof, omitting the major premiss, and giving the number of the proposion referred to as the authority for the minor premiss. II. At the point A, in the line A B, make an angle A B D H equal to CDE; and at the point B make an angle equal to FG H. PROBLEM (Euclid I. 2). Repeat.-The definition of a circle; and of an equilateral triangle: and Axiom 1. General Enunciation. From a given point to draw a straight line equal to a given straight line. Particular Enunciation. Given. The point A and the point A, a straight line B . A (NOTE.-The point A might be given in any required position. See Exercises on this proposition.) Construction. Join AC (see Fig. on next page). On A C describe the equilateral triangle D A C (Euc. I. 1). Produce the straight lines DA, DC, to E and F. From the centre C, with the radius CB describe the circle BGH, cutting C F in G. D From the centre D, with the radius DG, describe the circle G L K, cutting A E in L. K C The line A L shall be equal to BC. Proof. All lines drawn from the centre of a circle to the circumference are equal (Definition). C is the centre of circle BGH, and BC and CG are drawn from it to the circumference. (a).. BC is equal to C G. All lines drawn, etc. (definition of a circle). If equals be taken from equals the remainders are equal. Axiom 3. Take the equal lines DC, DA from the equal lines DC, DL. (c) Then the remainder C G is equal to the remainder AL. Lines equal to the same line are equal to one another. BC (a) and AL (c) are each equal to C G. ... AL is equal to B C. Result.-Wherefore from the given point A a straight line AL has been drawn equal to line B C. the given Q. E. F. A. EXERCISES.-I. Let A be the given point and BC the given line, then draw a line from A equal to BC. B D A (Join AC. Construct the equilateral triangle A CD. |