## Friday, May 4, 2012

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

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

Click the figure below to see the problem 748 details.

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

let 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

2. 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

1. To Peter, Problem 748, The extra information is for a short solution. Your solution is great because is more general, see problem 750.

3. Excellent solution by Peter!
Just wondering if there's any other way to determine x?

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

4. Quadrilateral ABCD is cyclic, from the two right angles so ptolemy's theorem applies

5. Apply Menelau's to triangle ABF
=> (AE/AB)*(BC/CF)*(FD/DA) =1
=> 7/4*4/3*(BC/CF)*7/4=1
=> BC/CF =12/49

As BCH & BFN are similar
x/7=12/61
=> x = 84/61

6. According to side perpendicular to each other
=> GEB=GBE=ABM=MBC=45° => GB=3, BM=4, BN=7 ??????