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

Labels:
angle bisector,
incenter,
proportions,
ratio,
triangle

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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