Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 632.
Thursday, July 7, 2011
Problem 632: Triangle, Interior and Exterior Angle Bisectors, Midpoints, Collinear Points
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Let AC=b , AB=c , CB=a and B’C/B’A=B”C/B”A= k (1)
ReplyDeleteFrom (1) we have B’C=k.b/(1+k) and B”C=k.b/(1-k)
CB”’=(CB”-CB’)/2 = b.k^2/(1-k^2)
And B”’A=B”’C+AC = b/(1-k^2)
So CB”’/B”’A= k^2= (a^2/c^2)
Similarly we also have C”’A/C”’B= b^2/a^2 and A”’B/A”’C=c^2/b^2
We have ( CB”’/B”’A).( C”’A/C”’B).( A”’B/A”’C) = a^2/c^2 . b^2/a^2 . c^2/b^2= 1
3 points A”’, B”’ and C”’ are collinear per Menelaus Theorem
Peter Tran