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.

## Monday, May 19, 2008

### Elearn Geometry Problem 79

Labels:
altitude,
incircle,
inradius,
orthocenter,
similarity,
triangle

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Tr. BFH is similar to Tr. CEH; hence c/f = CE/FB.

ReplyDeleteSimilarly, 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

Ajit: ajitathle@gmail.com

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.

ReplyDeleteQ. E. D.

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:

ReplyDeletec/f=CH/BH, b/e=BH/AH, and a/d=AH/CH, therefore:

(abc)/(def)=((CH)(BH)(AH))/((BH)(AH)(CH))=1

and the conclusion is the same but with one less theorem to memorize...