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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1219.
Tr.s BCC1 and BAA1 are congruent SAS so CC1 = AA1 hence from the mid point theorem, B1B2B4B3 is easily a rhombus with B1B4 and B2B3 perpendicular to each other
Balance....Let AA1 and CC1 meet at P. Since Tr.s BCC1 and BAA1 are congruent < BCC1 = < AA1B, BRCA1 is concyclic Hence < BRA1 = BCA1 = 60Similarly < BRC1 = 60Hence AA1 and CC1 make an angle 120 with each other and this therefore must be the case for B1B3 and B1B2 as well since these are respectively parallel to AA1 and CC1 by the midpoint theoremSince therefore B1B2B4B3 is a rhombus as proved earlier and < B2B1B3 = 120 easily B1B2B4 and B1B3B4 are equilateral trianglesSumith PeirisMoratuwaSri Lanka
Triangles ABA1 congruent to C1BC … ( case SAS)Triangle ABA1 is the image of C1BC in the rotation transformation centered B, rotation angle= 60So CC1= AA1 and angle formed by AA1 to CC1= 60 degrees.In triangle C1AA1, since B1, B4 are midpoint of AC1 and BC1 => B2B4 = ½. AA1 and B2B4// AA1Similarly B1B2= ½ CC1= B2B4 and B1B2// CC1 and angle formed by B2B4 to B1B2= 60 degreesSo B1B2B4 is equilateral. Similarly B1B3B4 is also equilateralSolution to all other questions will follow this result