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An Incentre Problem (Posted on 2007-04-02) Difficulty: 2 of 5
The incentre of triangle ABC is located at I. The incircle touches BC and CA respectively at D and E, and BI meets DE at G.

Prove that AG is perpendicular to BG.

  Submitted by K Sengupta    
Rating: 4.0000 (1 votes)
Solution: (Hide)
Angle CED = Angle CDE = 90o - Angle ACB/2
Hence, Angle AEG
= 180o - Angle DEC
= 90o + Angle ACB/2

Now, Angle AIG = Angle ABI + Angle BAI
= (Angle ABC + Angle BAC)/2

Hence, Angle AEG + Angle AIG
= 90o + (Angle ACB + Angle ABC + Angle BAC)/2
= 90o + 90o = 180o

Therefore, AEGI is a cyclic quadrilateral, so that:
Angle AGI = Angle AEI = 90o

Consequently, it follows that AG is perpendicular to BG.


An alternate methodology is posted by Bractals in this location.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionBractals2007-04-02 16:44:03
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