## Wednesday, June 25, 2008

### Elearn Geometry Problem 128

See complete Problem 128
Incenter of a Triangle, Angle Bisectors, Sum of Ratios. Level: High School, SAT Prep, College geometry

1. BI is a bisector of angle B
AI/ID = AB/BD
CI is a bisector of angle C
AI/ID = AC/CD
AI/ID = AB/BD = AC/CD
AI/ID = ( AB + AC )/( BD + CD )
AI/ID = ( AB + AC )/BC
( AI + ID )/ID = ( AB + BC + CA )/BC
ID/AD = BC/( AB + BC + CA )
Similarly
IE/BE = AC/( AB + BC + CA )
IF/CF = AB/( AB + BC + CA )
ID/AD + IE/BE + IF/CF = 1
Magdy Essafty

2. using theorem of van aubel we have :
AI\ID = AF\FB + AE\EC = AC\BC + AB\BC = (AC+AB)\BC
now AI\ID + 1 = AD\ID = s\BC (and s = AB + AC + BC)
and similar after sum we get : ID\AD + IE\BE + IF\FC = BC\s + AC\s + AB\s = s\s = 1 .
so we are done !!

3. To Anonymous:
As far as I know the theorem of Van Aubel, I can't see how it is applyed in your solution posted in April 13, 2012. Could you please explain it?
Thanks.

4. There is a simpler proof using areas:
= (IBC)/(ABC)
Similarly
IE/BE = (IAC)/(ABC) and
IF/CF = (IAB)/(ABC)
Hence