Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 709.
Thursday, December 29, 2011
Problem 709: Parallelogram, Midpoint, Diagonal, Metric Relations, Triangle
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http://img607.imageshack.us/img607/5235/problem709.png
ReplyDeleteExtend AF and BF to AD and BC( see picture)
Let AD=BC=DN=CL=2b
Let CD=2a, AF=FL=2c
Note that triangle AHD similar to triangle EHL so
AH/HL=AD/EL=HD/HE=2/3
From this ratio we get HF=2c/5 and AH/HF=4
Triangle BGE similar to triangle NGA so AG/GE=4b/b=4
So GH//EF//BD
Triangle HAD similar to triangle HLE
So AG/GE=AD/El=2/3 and EH/ED=3/5=3/x
So x=5
Trace EF. BD meet AE and AF at P and Q.
ReplyDeleteΔABQ ~ QFD --> BQ = 2QD
ΔAPD ~ EPB --> PD = 2BP
BD = BQ + QD = 3 QD --> QD = 1/3x
BD = BP + PD = 3 BP --> BP = 1/3x
By Thales’ theorem: EF // BD , EF = 1/2x.
ΔBPG ~ ΔGEF --> PG/GE=BP/EF =(x/3)/(x/2)= 2/3
ΔQHD ~ ΔHEF --> QH/HF=QD/EF =(x/3)/(x/2)= 2/3
Then PG/GE = QH/HF = 2/3 ----- ( 1 )
In the trapezoid EFQP, (1) --> GH // BD //EF or MH // BD //EF.
ΔMEH ~ ΔBED --> MH/BD = GE/EP = 3/(3+2) = 3/5
Then 3/x = 3/5 --> x=5.