Proposed Problem

Click the figure below to see the complete Routh's Theorem 4 about Triangle, Cevians, Ratio, Areas.

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Routh's Theorem 4

Level: High School, SAT Prep, College geometry

## Sunday, January 10, 2010

### Routh's Theorem 4: Triangle, Cevians, Ratio, Areas

Labels:
area,
Ceva's theorem,
cevian,
Menelaus' theorem,
ratio,
Routh's theorem,
triangle

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SB'A'C/SAA'B' = 1/m => SAA'C/SAA'B' = (1+m)/m

ReplyDeleteSAA'B' = km/(1+k)(1+m) (1) (SAA'C = k/(1+k) S from P1)

SAA'B' = AA'∙h3 ( h3 alt from B' to AA')

AA' = [km/(1+k)(1+m)]/[m/(1+m)] (h3 = m/(1+m) h1 P1

AA' = k/(1+k) h1 (2)

SAA'C' = AA'∙h4 ( h4 alt from C' on AA')

SAA'C' = AA'∙1/(k(1+n) h1 (h4 related to h from B and

h fro C see P1)

AA' = (SAA'C')/[1/(k(1+n)] (3)

from (2) and (3)

SAA'C' = 1/(1+k)(1+n) (4)

from (1) and (4)

SAB'A'C' = 1/(1+k)(1+n) + km/(1+k)(1+m)

SAB'A'C' =(1+m+kmn+km)/(1+k)(1+m)(1+n)

SB'C'A' = SAB'A'C' - SAB'C' (SAB'C'=m/(1+m)(1+n) P3

SB'C'A' = (1+m+kmn+km)/(1+k)(1+m)(1+n) - m/(1+m)(1+n)

SB'C'A' = (kmn+1)/ (1+k)(1+m)(1+n)

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second way, like P2

ReplyDeleteSA'B'C' = S - ( SA'BC' + SA'CB' + SAB'C' )

SA'B'C' = S - [n/(n+1)(k+1)+ k/(k+1)(m+1)+ m/(m+1)(n+1)]

SA'B'C' = (kmn + 1) / (k+1)(m+1)(n+1)

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