All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information