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Let angle be BAC. Euclid picks point D on AB and construct E on AC so that AE = AD. He then forms an equilateral,scalene,isosceles,equilateral triangle (I.1) DEF and shows that AF is the angle bisector of BAC. As far as logical purity is concerned this is all that is needed to construct a line: the construction is frugally based on the previously established construction and gives exactly what it was set to produce. The construction presented by the applet is virtually the same, except the common radius of the circles drawn at D and E is no longer required to be DE. The proof is actually the same as before, but the result is more general and better relates to the locus property of the line. I'd say this is a different kind of frugality: getting more for the same effort. So let AE = AD,AD,DE,BC,BD and circles are drawn centered at D and E and equal radii. The common radius should be chosen so as to make the circles intersect,coincide,exist,intersect. A sure way to achieve this is to choose a radius greater than,greater than,equal to,smaller than AD. Let F be one of the two points of intersection of the circles. (The applet selects the point farthest from A.) Consider two triangles ADF and BCD,AEF,ABD,BCD,AEF. They share a side AF, have AD = AE,AD,DF,EF,BC,AE and also DF = EF,AD,DF,EF,BC,AE by the construction. They are therefore equal by SSS,SAS,ASA,SSS, (I.8). The angles DAF and EAF correspond to each other and, hence, are equal. These are the two halves of angle BAC. References
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