tag:blogger.com,1999:blog-6933544261975483399.post4999775549435069115..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 709: Parallelogram, Midpoint, Diagonal, Metric Relations, TriangleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-29590623182058186832012-01-01T23:08:21.015-08:002012-01-01T23:08:21.015-08:00Trace EF. BD meet AE and AF at P and Q.
ΔABQ ~ QFD...Trace EF. BD meet AE and AF at P and Q.<br />ΔABQ ~ QFD --> BQ = 2QD <br />ΔAPD ~ EPB --> PD = 2BP<br />BD = BQ + QD = 3 QD --> QD = 1/3x <br />BD = BP + PD = 3 BP --> BP = 1/3x<br />By Thales’ theorem: EF // BD , EF = 1/2x.<br />ΔBPG ~ ΔGEF --> PG/GE=BP/EF =(x/3)/(x/2)= 2/3 <br />ΔQHD ~ ΔHEF --> QH/HF=QD/EF =(x/3)/(x/2)= 2/3 <br />Then PG/GE = QH/HF = 2/3 ----- ( 1 )<br />In the trapezoid EFQP, (1) --> GH // BD //EF or MH // BD //EF.<br />ΔMEH ~ ΔBED --> MH/BD = GE/EP = 3/(3+2) = 3/5<br />Then 3/x = 3/5 --> x=5.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65050756467632653392011-12-29T22:03:37.460-08:002011-12-29T22:03:37.460-08:00http://img607.imageshack.us/img607/5235/problem709...http://img607.imageshack.us/img607/5235/problem709.png<br /><br />Extend AF and BF to AD and BC( see picture)<br />Let AD=BC=DN=CL=2b<br />Let CD=2a, AF=FL=2c<br />Note that triangle AHD similar to triangle EHL so<br />AH/HL=AD/EL=HD/HE=2/3<br />From this ratio we get HF=2c/5 and AH/HF=4<br />Triangle BGE similar to triangle NGA so AG/GE=4b/b=4<br />So GH//EF//BD<br />Triangle HAD similar to triangle HLE <br />So AG/GE=AD/El=2/3 and EH/ED=3/5=3/x<br />So x=5Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com