perplexus.info :: Geometry : An Incentre Problem

An Incentre Problem (Posted on 2007-04-02)

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 = 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: ( You must be logged in to post comments.)

	Subject	Author	Date
	Solution	Bractals	2007-04-02 16:44:03

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