Geometry Problem
Click the figure below to see the complete problem 450 about Triangle, Median, Cevian, Congruence, Proportional Segments.
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Complete Problem 450
Level: High School, SAT Prep, College geometry
Tuesday, May 11, 2010
Problem 450: Triangle, Median, Cevian, Congruence, Proportional Segments
Labels:
cevian,
congruence,
isosceles,
median,
proportions,
similarity,
triangle
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draw CM//BD ( M on AE )
ReplyDelete▲BFE ~ ▲EMC
EM/y = (a - c)/c => EM = y(a-c)/c
F midpoint of AM =>
x = y + EM => x = y + y(a-c)/c => xc = yc + ay - yc
xc = ay
x/y = a/c
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by Menelaus' theorem
ReplyDeleteAF/FE*EB/BC*CD/AD=1 ⇔
x/y=a/c
use sine rule in ▲ABE and ▲FBE to get (sinABD/sinDBC)=(x/y) and then in ▲ABD and ▲CBD to get (sinABD/sinDBC)=(a/c)...equate them to get the result...
ReplyDeleteProblem 450
ReplyDeleteABC is the triangle area (ABC).Τhen (ABF)/(BEF)=x/y=(ADF)/(DFE)=(ABD)/(DBE)=υ_1/υ_2.
(Where υ_1and υ_2 are the heights of the triangles ABD, BED, which correspond to the sides AB, BE).But 1=(ABD)/(BCD)=(c. υ_1)/(a. υ_2) so υ_1/υ_2=a/c.Therefore a/c=x/y.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Draw CG //BD where G is on AE extended
ReplyDeleteTr.s ECG and BFE are similar
Hence (x-y)/y = (a-c) / c
So x/y = a/c
Sumith Peiris
Moratuwa
Sri Lanka
Draw segment ME//AC:M on AB and intersecting BD at G.
ReplyDeleteConsider Tr.FEG and Tr.AFD: they are similar by AAA.
y/x=EG/AD=FG/FD.
Consider Tr.BGE and Tr. BDC: they are similar by AAA.
EG/AD=EG/DC since AD=DC(given) but EG/DC=BE/DC=c/a.
Hence, y/x=c/a