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.

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