## Tuesday, May 11, 2010

### Problem 450: Triangle, Median, Cevian, Congruence, Proportional Segments

Geometry Problem
Click the figure below to see the complete problem 450 about Triangle, Median, Cevian, Congruence, Proportional Segments.

Complete Problem 450
Level: High School, SAT Prep, College geometry

1. draw CM//BD ( M on AE )
▲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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2. by Menelaus' theorem
x/y=a/c

3. 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...

4. Problem 450
ABC 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

5. Draw CG //BD where G is on AE extended

Tr.s ECG and BFE are similar

Hence (x-y)/y = (a-c) / c

So x/y = a/c

Sumith Peiris
Moratuwa
Sri Lanka

6. Draw segment ME//AC:M on AB and intersecting BD at G.
Consider Tr.FEG and Tr.AFD: they are similar by AAA.