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An Incentre Problem (
Posted on 20070402
)
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:
(
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)
Angle CED = Angle CDE = 90
^{o}
 Angle ACB/2
Hence, Angle AEG
= 180
^{o}
 Angle DEC
= 90
^{o}
+ Angle ACB/2
Now, Angle AIG = Angle ABI + Angle BAI
= (Angle ABC + Angle BAC)/2
Hence, Angle AEG + Angle AIG
= 90
^{o}
+ (Angle ACB + Angle ABC + Angle BAC)/2
= 90
^{o}
+ 90
^{o}
= 180
^{o}
Therefore, AEGI is a cyclic quadrilateral, so that:
Angle AGI = Angle AEI = 90
^{o}
Consequently, it follows that AG is perpendicular to BG.
******************************************
An alternate methodology is posted by Bractals in this
location
.
Comments: (
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Subject
Author
Date
Solution
Bractals
20070402 16:44:03
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