Saturday, January 15, 2011

Problem 567: Right triangle, Incircle, Excircle, Collinear tangency points

Geometry Problem
Click the figure below to see the complete problem 567.

Problem 567: Right triangle, Incircle, Excircle, Collinear tangency points.
Zoom at: Geometry Problem 567

Posted by Antonio Gutierrez at 5:41 AM
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4 comments:

  1. Peter TranJanuary 15, 2011 at 9:28 AM

    http://img843.imageshack.us/img843/4514/problem567.png

    Let 2p=perimeter of tri. ABC ; r= radius of incenter circle
    We have CF=p and AG= p-BC
    FB=CF-BC=p-BC
    So AG=FB
    Triangle FBG congruent to tri. AGD ( case SAS)
    and angle(FGB)=angle(ADG)=angle(AGH)
    So F, G, H are collinear

    Peter Tran

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  2. c .t . e. oJanuary 15, 2011 at 9:50 AM

    EA//GH ( EAH + GHA = 90 + A/2 + 90 - A/2 )
    EF//AG, EF = AG = R => EFGA paralelogram
    => EA//FG
    from G two // to AE => F, G, H collinear

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  3. André SilvaJanuary 24, 2011 at 6:57 PM

    Other,
    ADGH são cocíclicos => DGH=a/2 (i)
    M é a interseção de AD com GH, então AMBF tb é cocíclico e BFG = a/2 (ii)
    De (i) e (ii) BGF = 90 - a/2 = AGH => F, G, H colineares.

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  4. APOSTOLIS MANOLOUDISJune 22, 2016 at 2:16 PM

    If the circle centered on the point D intersects the BC in K and s is half-perimeter of triangle ABC.(AH=AG=s-a=BF, CH=CK=s-c, BG=BK=s-b).Then CH/HA.AG/GB.BK/KC=1 and from the inverse theorem Menelaus, the points F,Gand H are collinear.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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