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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 1232.
https://goo.gl/photos/1o6iZtQfZnuheshYAConsider complete quadrilateral ABC C1A1B2Diagonals AA1 and CC1 meet at HSo ( ACB1B2)= -1 or B1, B2 are harmonic conjugate points of A and C…..Or B2A/B2C=B1A/B1C…. (1)Similarly we also have A2C/A2B= A1C/A1B…… (2)And C2B/C2A=C1B/C1A….. (1)Multiply expressions (1), (2) and (3) side by sideThe value of right hand side is 1 per Ceva theorem So the left hand side B2A/B2C x A2C/A2B x C2B/C2A= 1And A2, C2, B2 are collinear per inverse of Menelaus theoremNote: This property can be generalized for any lines AA1, BB1, CC1 as long as that these lines concur at a point
Easily A1A2, B1B2 and C1C2 are external angle bisectors of the Orthic Triangle A1B1C1These angle bisectors meet the opposite sides at A2, B2 and C2From the result of Problem 631 these points are collinear Sumith PeirisMoratuwaSri Lanka