Tuesday, December 11, 2018

Geometry Problem 1408: Right Triangle, Incircle, Excircle, Incenter, Midpoint, Tangency Point, Collinearity

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1408: Right Triangle, Incircle, Excircle, Incenter, Midpoint, Tangency Point, Collinearity, Tutoring.

Posted by Antonio Gutierrez at 4:48 PM
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2 comments:

  1. AnonymousDecember 11, 2018 at 8:13 PM

    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.


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  2. Sumith PeirisDecember 15, 2018 at 1:09 AM

    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,

    XF/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

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