
Re: Three points on incircle
Posted:
Apr 28, 2014 1:04 PM


> Let ABC be an equilateral triangle with side length a > = 2. Construct points D, E, F on the incircle, such > that triples {C,D,E}, {B,E,F}, {A,F,D} are collinear > respectively. > > > Best regards, > Avni Hi Avni,
Let incenter = I Bisect BI at H and construct circle on BI as diameter Bisect HI at J and construct circle, center I, radius IJ to intersect first circle at G. Then G is the midpoint of EF and B,E,F are collinear sin(angleIBG)=((inradius/2)/circumradius)=1/4 So angle between side BA and side EF =Pi/6arcsin(1/4) =3014.48degrees=15.52deg approx.
Regards, Peter Scales.

