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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.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Solution Solution Comment 1 of 1


We will show that <AGI is congruent to <AEI = 90.
If G = E, then <AGI is <AEI; otherwise,
let x, y, and z be the measures of the
half-angles of A, B, and C respectively.
Clearly, x+y+z = 90. In triangle BGD we have
  <BGD = 180 - <GDB - <DBG
       = 180 - <GDI - <IDB - <DBG
       = 180 - z - 90 - y
       = x
If E lies between D and G, then
  <IAE = <IGE = <BGD = x
  AGEI is cyclic since side IE subtends equal
  angles at the opposite vertices.
;otherwise 
  <BGD is an exterior angle of triangle IGE.
  <BGD = <IEG + <GIE
  <AEG + <GIE = (<AEI + <IEG) + (<GIE + <EIA)
              = <AEI + <BGD + <EIA
              = 90 + x + (90 - x)
              = 180
  AEGI is cyclic since opposite angles are
  supplementary.
Therefore,
  <AGI = <AEI as angles subtended by side AI.

 

Edited on April 2, 2007, 4:47 pm
  Posted by Bractals on 2007-04-02 16:44:03

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