Particular Enunciation. Given.—The straight line AB and the point C in AB Required.–To draw from the point ca straight line at right angles to A B. Construction. (a) In A C take any point D. (6) From CB cut off a part CE equal to CD. (Euc. I. 3.) (c) On D E describe the equilateral triangle DFE. (Euc. I. 1.) (d) Join FC. The straight line FC drawn from the given point C, shall be at right angles to the given straight line A B. If FC be at right angles to A B, we must prove that angle FCD is equal to angle FCE. Proof (with Syllogisms in full). D C equals CE (by Construction 6); add to each CF. (e) Then by Axiom 2a, DC, CF are equal to EC, CF, each to each. Ist Syllogism. If two triangles have two sides of the one equal to two sides of the other, and the base of the one equal to the base of the other; those sides shall contain equal angles (Euc. I. 8). The two triangles FCD, FCE have the two sides D C, C F equal to the two sides EC, CF (e), each to each, and they have the base D equal to the base E F (by construction (c), FDC being an equilateral triangle). (f).. the angle DCF is equal to the angle ECF. 2nd Syllogism. When a straight line, standing on another straight line, makes the adjacent angles equal, each of the angles is called a right angle (Euc. Def. 10). The straight line F C standing on the straight line AB makes the angle DCF equal to the adjacent angle ECE (f). .. Each of the angles DCF, ÉCF is called a right angle. Result.-Wherefore from the given point C, a line CF has been drawn at right angles to A B. C.E.F. EXERCISES.—I. Write out this proof in contracted syllogisms. II. At the point N in the given N straight line N O draw a straight line at right angles to NO (produce N O towards N). III. At the point N in the straight line given in (II), draw a straight line double the length of N O at right angles to No. (Produce N O both ways; make the produced parts each equal to No; draw a line at right angles from N; describe a circle with N as centre and radius twice N0; produce line at right angles to NO till it meets the circle.) PROBLEM (Euclid I. 12). Repeat.-The definition of a circle, and of a right angle: and the enunciation of Euc. I. 8 (page 46). General Enunciation. To draw a straight line perpendicular to a given straight line from a given point without it. A B and the point C. line perpendicular to A B. Construction. Take any point D on the side of A B remote from C At the centre C and distance CD describe the circle FDG, cutting A B in FG. (The whole circle is not shown in the figure, only the part needed.) (a) Bisect FG (Euc. I. Io shows how), and call the point of bisection H. Join CH The straight line CH drawn from the given point Cis perpendicular to the given straight line A B. If CH is perpendicular to A B, the angle CHF will be equal to the angle CHG. Then we must prove that angle CHF is equal to angle CHG. Proof (with syllogisms in full). Additional construction required for the purposes or proof. Join CF, CG ist Syllogism. All lines drawn from the centre of a circle to the circumference are equal. (Definition of a circle.) CF and C G are drawn from the centre C to the circumference FDG. (6) ..CF is equal to CG. FH is equal to HG (by Construction a). (C).. FH, H Care equal to G H, H C, each to each. 2nd Syllogism. equal to two sides of the other, each to each, contain equal angles. (Euc. I. 8.) sides FH, HC equal to the two sides base CHF (6). angle CHG line makes the adjacent angles equal to (definition). equal to the adjacent angle CHG (d). .. C H is perpendicular to A B. Result.-Wherefore a perpendicular CH has been drawn to the given line A B from the given point C. Q. E. F. EXERCISES.—I. Write out the proof of this proposition, omitting the major premiss of each syllogism, and giving the definition or proposition of Euclid referred to as the authority for the minor premiss. [Thus the first syllogism will read: Because C F and C G are drawn from the centre C to the circumference FDG... CF is equal to CG. (Euc. Def. of circle.)] |