Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 748 details.

## Friday, May 4, 2012

### Problem 748: Quadrilateral, 90 Degrees, Diagonal, Perpendicular, Metric Relations

Labels:
90,
degree,
diagonal,
metric relations,
perpendicular,
quadrilateral

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http://img859.imageshack.us/img859/6765/problem748.png

ReplyDeletelet 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= 3/4

In triangle ∆AEF, applying Ceva’s theorem => DA/DF x PF/PE x BE/BA=1

Replace DA/DF=4/7 and BE/BA=3/7 we get PF/PE=7/3

In triangle AEF, applying Von Aubel’s theorem => CF/CB=PF/PE+DF/DA

Replace PF/PE=7/3 and DF/DA=7/4 we get CF/CB=49/12

So CB/BF=12/61=x/7

X=84/61

By looking my solution as above, I found that angles B and D need not be 90 degrees.

ReplyDeleteI 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.

DeleteExcellent solution by Peter!

ReplyDeleteJust wondering if there's any other way to determine x?

To Ajit, problem 748, yes, there is another way to determine x.

Delete