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