See complete Problem 128
Incenter of a Triangle, Angle Bisectors, Sum of Ratios. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Wednesday, June 25, 2008
Elearn Geometry Problem 128
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BI is a bisector of angle B
ReplyDeleteAI/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 )
Therefore by addition
ID/AD + IE/BE + IF/CF = 1
Magdy Essafty
using theorem of van aubel we have :
ReplyDeleteAI\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)
so : ID\AD = BC\s
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 !!
adil azrou from morocco .
To Anonymous:
ReplyDeleteAs 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.
There is a simpler proof using areas:
ReplyDeleteID/AD
= (IDB)/(ADB)= (IDC)/(ADC)
= [(IDB)+(IDC)]/[(ADB)+(ADC)]=(IBC)/(ABC)
= (IBC)/(ABC)
Similarly
IE/BE = (IAC)/(ABC) and
IF/CF = (IAB)/(ABC)
Hence
ID/AD + IE/BE + IF/CF
=[(IBC)+(IAC)+(IAB)]/(ABC)=ABC)/(ABC)
= 1
Note:In fact PD/AD + PE/BE + PF/CF = 1
holds for any three concurrent cevians APD, BPE, CPF
To Anonymous: My previous question is already solved. I've learned more about Van Aubel. Thanks.
ReplyDelete