Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 634.
Friday, July 15, 2011
Problem 634: Right Triangle, Altitude, Incenters, Areas
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Note that Triangles ABC, BHC and ABH are similar triangles ( Case AA)
ReplyDeleteIn each triangle , ratio of height from 90 degrees angle/ radius of incenter circle = h1/r1= h2/r2=h3/r3
So ratio S(AEB)/S(AHB)=S(BFC)/S(BHC)=S(ADC)/S(ABC)
But we have S(ABC)=S(ABH)+S(BHC)
So S(AEB)+S(BFC)=S(ABC)
Peter Tran
Call α = angle A, AB = c, BC = a and AC = b.
ReplyDeleteBecause AE, BF and CD are bisectors, it is easy to see that ∆AEB, ∆BFC and ∆ACD all have angles ½α, 45º-½α and 135º, so they are all similar triangles.
factor ∆ACD → ∆ABE = c/b so S1 = S3*(c/b)².
factor ∆ACD → ∆BCF = a/b so S2 = S3*(a/b)².
S1 + S2 = S3*((c/b)² + (a/b)²) = S3*(a²+c²)/b²
Pythagoras in ∆ABC: a²+c²=b²
S1 + S2 = S3*b²/b² = S3.
QED.
(Note: there is a little error in the text on the side of the picture. F is the incenter of ∆BHC.)
Thanks Henkie for the note.
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