let incircle radius r, touch BC at H; excircle radius R, Right ABC, the length is a,b,c, c>a,c>b; r=(a+b-c)/2; R=(c+a-b)/2; if triangle IHF similar to triangle BDF, then D,I F colinear, it easy to show r/(b/2) =(R-r)/R so D,I,F colinear.
If X & Y are the tangency points of the in circle at BC & BA & if the radii of the incircle is p & of the excircle is q & S(ABC) = S, s the semi perimeter,
let incircle radius r, touch BC at H;
ReplyDeleteexcircle radius R,
Right ABC, the length is a,b,c, c>a,c>b;
r=(a+b-c)/2; R=(c+a-b)/2;
if triangle IHF similar to triangle BDF, then D,I F colinear,
it easy to show r/(b/2) =(R-r)/R
so D,I,F colinear.
If X & Y are the tangency points of the in circle at BC & BA & if the radii of the incircle is p & of the excircle is q & S(ABC) = S, s the semi perimeter,
ReplyDeleteXF/IX = (q-p)/p = (a-2p) / p = (a-2S/s)/(S/s)
= (as - ac)/(ac/2) = (s-c)/(c/2) = BF/BD
So Tr.s BDF & XIF are similar and DIF are collinear
Sumith Peiris
Moratuwa
Sri Lanka