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 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.

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

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 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 thehalf-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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