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
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.
<BGD is an exterior angle of triangle IGE.
<BGD = <IEG + <GIE
<AEG + <GIE = (<AEI + <IEG) + (<GIE + <EIA)
= <AEI + <BGD + <EIA
= 90 + x + (90 - x)
AEGI is cyclic since opposite angles are
<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