> Dear Sujeet, > > Here is my solution. Without trig of course... > > Construct isosceles triangle APC with base angle PCA > = PAC = 10 deg, > hence with APC angle = 160 deg. > Construct equilateral triangle PCD, of course PD = > AP. > Draw circumcircle of APD. > Because the circumcenter is on the perpendicular > bisector of PD, > a diameter B'C of this circle is the angle bisector > of PCD > > Angle DPA = 360 - 60 - 160 = 140 deg, of which we > deduce at once > angle DPO = 140/2 = 70 deg, hence DOP = 180 - 70 - 70 > =) 40 deg > hence DB'P = 20 deg and CB'P = 10 deg > > But what relation with our problem ??? > Here it is : > angle DAB' = angle DPB' = (180 - 20)/2 = 80 deg > Angle DAP = angle DB'P = 20 deg > Hence PAB' = 100 deg and that's it ! > the point B' in the figure is the same as point B in > the problem ! > > Regards.
Thank you, for showing new way to attack the problem.I always thought this may be a way to give solution, but I never got success. In 2002, in a similar kind of discussion , I came across following problem- Q.The Triangle ABC is isosceles with equal sides AC and BC .Two of its angles measures 40 degree, the interior point M is such that angle MAB =10 degree and angle MBA=20 degree.Determine the measure of angle CMB. Author had mentioned that it was a good problem acording to his point of view because it had destroyed his self confidence in geometry. That time for me this was a new kind of problem, where a point is taken inside the triangle.Just 6 months ago I had found that isosceles triangles with vertical angles 20 degree , 100 degree and 140 degree have similar kind of relationship among their sides and angle bisectors.I used those properties to solve above mentioned problem.It made me very excited about these kind of problems. Again , later I found another discussion , which was on this site only, dated 3 September, 2001.But I had seen this in early 2003.Here Dave Rusin ,again mentioned this problem , he had not given proof but he had given a list of all such problems where angles are multiple of 10 degree.There were approximately 75 such combinations where all 6 angles are multiples of 10 degree.One can form a problem by giving 4 angles and task is to find remaining ones.From that list, I took first 4 to be known and last 2 to be found out.Among 75 , 35 are trivial or too easy. Remaining 40 are not that easy to prove.From 40 , till now I have proved only 16.24 problems are still resisting pure geometric proof.I have already posted most of them , in coming few days it will be over. In same discussion , Dave rusin posted another list where angles are multiples of 6 , 7.5 and 15 degree. I have never tried those problems where angles are mutiples of angles other than 10 degree.I think I will try those when problems with angles multiples of 10 degree are over. In same dicussion Ayatollah Potassium had written that there is one advantage of geometric approach over other methods-if you verify that a given solution works , but the proof doesn't use all the information in the diagram, then there is a family of solutions including the given one. This has been the situation in most of the 16 problems I have solved.Only 2 problems needed properties of 3 triangles I have above mentioned., to be solved.In other problems , angles need not be multiples of 10 degree.There I replaced the angle with x. Another related thing I found in The Mathematical Intelligencer Vol. 17 No-1, 1995, titled ?19 Problems on Elementary Geometry? authored by Armando Machado.He started the article with Langley Problem , he had similar experiece what everybody has when he first tries to prove it. He also gave another list , this time it was of adventitious quadrangles, using software Mathematica. But these quadrangles had restriction that 2 sides extended leads to isosceles triangle.Now coming to main thing I want is that-Any interested reader who has got softwre Mathematica , help me to get complete list of adventitious quadrangles, specially quadrangles with angles multiples of 10 degree. Another thing is that Armando Machado had mentioned that in his list some problems showed duality He had used Pappus theorem from projective geometry.My knowledge of projective geometry is not very praise worthy.So please some one tell me how projective geometry can be used in these problems. What is duality?How 2 problems can be dual of each other? Regards , Sujeet