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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 748 details.
http://img859.imageshack.us/img859/6765/problem748.pnglet AC cut EF at P ( see sketch)∆DMA similar to ∆DNF so DA/DF=MA/FN=4/7∆BMA similar to ∆BGE so BE/BA=GE/AM= 3/4In triangle ∆AEF, applying Ceva’s theorem => DA/DF x PF/PE x BE/BA=1Replace DA/DF=4/7 and BE/BA=3/7 we get PF/PE=7/3In triangle AEF, applying Von Aubel’s theorem => CF/CB=PF/PE+DF/DAReplace PF/PE=7/3 and DF/DA=7/4 we get CF/CB=49/12So CB/BF=12/61=x/7X=84/61
By looking my solution as above, I found that angles B and D need not be 90 degrees.I am not sure whether I may miss something in my solution or author give us extra information which is not used in the solution. Please clarify.Peter Tran
To Peter, Problem 748, The extra information is for a short solution. Your solution is great because is more general, see problem 750.
Excellent solution by Peter!Just wondering if there's any other way to determine x?
To Ajit, problem 748, yes, there is another way to determine x.