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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 1026.
http://s25.postimg.org/t1ngf6pmn/pro_1026.pngDraw circle with center at B, radius= BEDraw altitude BM of triangle ABC. CE cut DL at HWe have ∠(ABL)= ∠(LBM)= ∠(MBD)= α And BL ⊥CHTriangle BCL is isosceles => BH= 1/2BL= 1/2BEIn triangle BHE since BH= 1/2BE => ∠(HBE)=60And x= ½∠(HBE)= 30Note that value of x doesn’t not depend on α
Let BF be the altitude of Tr. ABC and let BGH be the bisector of < ABF with G on CE and H on AC. Since BF the altitude of Tr. DHB, BH = BD = BE = HE since CEG the angle bisector of < C is also perpendicular to BGH. So BHE is equilateral and hence B is the incentre of Tr. HDE in which arc HE subtends 60 degrees at the centre B. Hence HE will subtend an angle of 30 degrees on the rest of the circle which shows that x the extrerior angle must be 30 Sumith PeirisMoratuwaSri Lanka
B is the circumcentre of Tr. HDE and not incentre
Or alternatively as per my above proof if CE meets BF, BH and BA at P,Q,R BPHR is easily seen to be a rhombus BE = 2 BQ and so < REH = 30 and therefore x = 30 from angle chasing (90 -2@+3@ = x+x+@)