Saturday, April 4, 2009

Problem 20: Right Triangle, Altitude, Inradii, Perpendicular

Proposed Problem

Problem 20: Right Triangle, Altitude, Inradii.

See complete Problem 280 at:
http://gogeometry.com/problem/p020_right_triangle_incircle.htm

Level: High School, SAT Prep, College geometry

3 comments:

  1. Using similarity between various rt. triangles that are there in the given diagram, it can be shown that: AD=AB^2/AC,CD=BC^2/AC, AE=AB^3/AC^2, DE=AB^2*BC/AC^2,DF=(AB*BC/AC)^2 and EF=BC*AB/AC.
    Using these relations and the fact that inradius = Tr.Area/Semisum of sides, we can show that:
    a = BC*AB^3/(AB+BC+CA)AC^2
    Likewise, b = AB*BC^3/(AB+BC+CA)AC^2
    while x =(BC*AB/AC)^2*(1/(AB+BC+CA)
    From these relations we can conclude that x^2=a*b
    QED
    Ajit: ajitathle@gmail.com

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  2. ahmetelmas-geo-geo-antonio.blogspot.com/

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  3. All triangles AED, EDF and DFC are similar and we have
    AE/ED=a/x
    AE/DF=AE/EB=a/b
    In right triangle ADB we have ED^2=AE.EB so x^2=a.b

    Peter Tran

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