See complete Problem 127

Triangle, Centroid, Incenter, Parallel, Proportions. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, June 23, 2008

### Elearn Geometry Problem 127

Labels:
angle bisector,
centroid,
incenter,
incircle,
median,
parallel,
proportions,
triangle

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1)[AGC]=[ABC]/3=sr/3=br/2 hence b=(a+c)/2

ReplyDelete2)M midpoint of AC; bisector BI and AC meet at D

AM=b/2; DC= ab/(a+c)=a/2

triangles BGI and BMD are similar with the ratio

2/3

GI=(2/3)MD= (c-a)/6

Solution to problem 127.

ReplyDeleteLet be BS the angle bisector and BM the median. From problem 126 we know that BI/IS = (a+c)/AC. Since GI and MS are parallel, then

BI/IS = BG/GM = 2,

so (a+c)/AC = 2 and AC = (a+c)/2.

As BS is the angle bisector, we have

c/AS = a/SC = (c-a)/(AS-SC) = (c+a)/(AS+SC) = (c+a)/AC = 2.

Thus (c-a)/(AS-SC) = 2 (1).

Besides, AS – SC = AM + MS – SC = MC + MS – SC = MS + SC + MS – SC = 2MS.

As BGI and BMS are similar, MS = (3GI)/2, then AS – SC = 2.(3GI)/2 = 3GI.

From (1) we get (c-a)/(3GI) = 2 and GI = (c-1)/6.

(GAC) = (IAC)

ReplyDelete∆/3 = (1/2)(br)= b∆/2s

3b = 2s

3b = a + b + c

b = (a + c)/2

Extend GI either way to meet BA at D and BC at E.

DG = GE = DE/2 = 2AC/6 = b/3 and BI/IE = sin C / sin (B/2)

GI = GE - IE = b/3 - BI sin(B/2)/sin C = b/3 - r/sin C

= b/3 - ra/h where h is altitude from B

But r = h/3 since GI∥AC

So GI = b/3 - a/3 = (b- a)/3 = (2b - 2a)/6 = (c - a)/6