Monday, May 19, 2008

Elearn Geometry Problem 79

See complete Problem 79
Triangle Similarity, Altitudes, Orthocenter, Incircles, Inradii. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. Tr. BFH is similar to Tr. CEH; hence c/f = CE/FB.
    Similarly, b/e = BD/EA & a/d = AF/DC. This gives us,abc/def = AF/DC * BD/EA * CE/FB
    or abc/def = AF/FB * BD/DC * CF/FA
    = 1 by Ceva's Theorem aince AD, BE & CF are concurrent in H.
    Thus, abc = def

  2. notice that if two triangles are similar then the ratio of their inradii equals the ratio of the corresponding sides. if Δ denotes the area of a triangle and Δ' is the area of another triangle similar to it then if r and r' be their inradii and if s and s' be their semi perimeters and let a and a' be any two sides of the triangles. we then have r:r' = (Δ/s):(Δ'/s') = Δ/Δ'*s'/s = a^2/a'^2*a'/a = a:a'. now in thw given figure triangles AFH and CDH are similar because angle AFH = angle CDH = 90 degrees and angle FAH = angle DCH as they are in he same segment. so AF/DC = a/d. similarly we have b/e = BD/AE and c/f = EC/BF. multiplying these relations we get a.b.c/d.e.f = AF.BD.EC/DC.AE.BF but as the altitudes are concurrent we have by ceva's theorem AF.BD.EC = DC.AE.BF. so AF.BD.EC/DC.AE.BF = 1 and therefore a.b.c/d.e.f = 1 which implies that a.b.c = d.e.f.
    Q. E. D.

  3. Joe's comment is right, but it can be somewhat shorter if we select the ratio of the longest sides instead of the ratio shown:
    c/f=CH/BH, b/e=BH/AH, and a/d=AH/CH, therefore:
    and the conclusion is the same but with one less theorem to memorize...