## Monday, June 16, 2008

### Elearn Geometry Problem 124 See complete Problem 124
Area of triangle, Similarity, Trisection of sides. Level: High School, SAT Prep, College geometry

1. AH/HB" * 1/3 * 2/1 = 1; hence AH/HB" = 3/2
BD/DB" * 1/3 * 2/1 = 1; hence BD/DB" = 3/2
Or HD is parallel to AC & HD/BB" = AH/(AH+HB")= 3/5 or HD/AC = 3/15 = 1/5
Similarly HF is parallel to BC and HF/BC = 1/5
and FD is parallel to AB and FD/AB = 1/5
Thus, Tr.HFD is similar to Tr.ABC and therefore their areas are proportional to the squares of their sides, so Tr. HFD/ Tr. ABC =(1/5)^2 = 1/25 or S1 = S/25
Ajit: ajitathle@gmail.com

2. name S point AH meet C'F

SASF = 2/15 - 1/21 ( SASC' = 1/7 ∙S/3 , 1/7 is C'S )

SASF = 3/35

using triangles ASF and SFH ( like P123 row 3-5 )

SSFH = 6/175 ( SASF = 3/35,AS = 3/7 AA',SH = 6/35AA')

SAHF = 3/35 + 6/175

SAHF = 21/175

for three sides

3∙21/175 = 63/175 (1)

SAFC = 1/5 ( from P122,123 )

for three sides

3∙ 1/5 = 3/5 (2)

3 (SAHF + SAFC)= 63/175 + 3/5

3 (SAHF + SAFC) = 168/175

S▲ = 175/175 - 168/175

S▲ = 1/25

3. Solution to problem 124.
Let K be the center of gravity, and AL, BN, CP the medians of triangle ABC. Let S be the area of ABC. As proved in problem 121,
KD/KL = KF/KN = KH/KP = 2/5.
So triangles KDF and KLN are similar with factor 2/5.
Then S(KDF) = (4/25).S(KLN).
Similarly, S(KFH) = (4/25).S(KNP),
and S(KHD) = (4/25).S(KPL). Adding these three equalities we get S1 = (4/25).(S/4) = S/25.