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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s12.postimg.org/a4sui5wi5/pro_1118.pngSince B3C4 and CC4 bisect angles B1B3C and B3CB1 ( see problem 1116)So B1C4 will bisect angle B3B1CAnd ∠ (B3B1C4)= ½ . 60= 30 degreesAnd ∠ (C3B1C4)= ∠ ( C3B3C4)= 90 => B3, C3, B1, C4 are concyclicSimilarly we also have B1A4 bisect angle AB1B3 and points A3, B3, A4 ,B1 are concyclic
Kindly refer my proofs for Problems 1114 thro to 1117. We have seen that < C3B1A4 = A4B1B2=< B2B1C4 = C4B1A3 = 30 Hence < C3B1C4 is 90 and since < C3B3C4 = 90 C3B3C4B1 is cyclicSimilarly A4B3A3B1 is cyclic Sumith PeirisMoratuwaSri Lanka
Another problem arising from this- Prove B1B2A4C3 is cyclic and so is B1A3C4B2Sumith PeirisMoratuwaSri Lanka