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

Labels:
diagonal,
metric relations,
midpoint,
parallelogram

Subscribe to:
Post Comments (Atom)

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.