General Enunciation. To describe an equilateral triangle on a given finite straight line. (A finite line is one whose length is definitely given.) (8) From the centre B, at the distance A B, describe the circle ACE. (c) From the point C, at which the circles cut one another, draw the lines CA, CB, to A and B. C A B E : Then A CB shall be an equilateral triangle (now we must prove that A B C is equilateral). Proof (with complete syllogisms). Ist Syllogism. All lines drawn from the centre of a circle to the circumference are equal (Definition of a circle). A is the centre of a circle, and A B and A Care drawn from it to the circumference (Const.). (a) ... A B is equal to AC (1st conclusion). 2nd Syllogism. All lines drawn from the centre of a circle to the circumference are equal (Definition of a circle). B is the centre of a circle, and A B and B Care drawn from it to the circumference. (b)... A B is equal to B C (2nd conclusion). 3rd Syllogism. Things equal to the same thing are equal to one another (Axiom I). A C and B C are each equal to A B (a and b). ... AC is equal to BC. Result... AC, BC, AB are all equal to one another. Wherefore the triangle A B C is equilateral, and it is described on the given straight line A B. Q. E. F. Proof (with contracted syllogisms, in the words of Euclid). Because A is the centre of the circle B CD. (a).. A C is equal to A B. Because B is the centre .. B C is equal to A B. But it has been proved A B (a). (Definition of circle.) of the circle A CF. (Definition of circle.) that A C is equal to (b) A C and B Care each equal to A B. Things which are equal to the same thing are equal to one another (Axiom 1). (c) ... A C is equal to B C. .. АС, ВС, AB are equal to one another (b, c). Wherefore the triangle A B C is equilateral, and it is described on the given straight line A B. Q. E. F. EXERCISES.-I. Describe an equilateral triangle having each of its sides double the given straight line OP. 0 P II. Describe two equilateral triangles, one on each 2 and side of the given line I prove that the five sides are all equal (use figures instead of letters to name the lines and points). PROBLEM (Euclid I. 22). Repeat. The definitions of a circle and a triangle. And Axiom 1. General Enunciation. To make a triangle, of which the sides shall be equal to three given lines. (Of the three given lines, any two must always be greater than the third.) Required. To make a triangle, whose sides shall be equal to A B and C. Construction. (a) Draw a straight line DE, of unlimited length Cut off DFequal to line A; FG equal to ine B; GH equal to line C. (Euc. I. 3 shows how to do this.) b) From Fas centre, with the radius FD, describe the circle DKL. (c) From Gas centre, with radius GH, describe the circle HKL, cutting the former circle at K. (d) Join FK, K G. The triangle shall have its sides FK equal to A; FG equal to B; GK equal to C. We are required to prove three things: 1. That FK is equal to A. 2. That FG is equal to В. 3. That GK is equal to C. Proof (with complete syllogisms) of No. 1. All lines drawn from the centre of a circle to the circumference are equal (Definition of circle). Fis the centre of circle DKL, and FK, FD are drawn from it to the circumference (Const.). (a) .. F K is equal to FD. (1st conclusion.) 2nd Syllogism. (a) FK is equal to FD. (1st conclusion become A is equal to FD (by construction). Proof (with complete syllogisms) of No. 3. Ist Syllogism. All lines drawn from the centre of a circle to the circumference are equal. (Definition.) Gis the centre of circle HKL, and GK, GH are drawn from it to the circumference. (a) .: G H is equal to GK. (1st conclusion. 2nd Syllogism. (a) GH is equal to GK. (Ist conclusion become a premiss.) GH is equal to C (by construction). (3) .: G K is equal to C. (Axiom 1.) (3rd conclusion wanted.) Proof of No. 2. (2) FG was made equal to B. Result. Collecting together conclusions (1), (2), (3), we have : The three sides KF, FG, GK are equal to the three lines A, B, C. |