Geometry Problem
Click the figure below to see the complete problem 545 about Acute Triangle, Squares, Altitudes, Area.
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Thursday, December 2, 2010
Problem 545: Acute Triangle, Squares, Altitudes, Area
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AB² - AB∙BF = AC∙AE see P 521
ReplyDeleteAC² - AC∙AE = BC∙DC
BC² - BC∙DC = AB∙BF
=>
AB² + AC² + BC² = 2∙( AB∙BF + AC∙AE + BC∙DC )
Apply cosine formula in triangle ABC we have:
ReplyDelete2.b.c.Cos(A)=b^2+c^2-a^2
2.a.c.Cos(B)=a^2+c^2-b^2
2.a.b.Cos(C)=a^2+b^2-c^2
add both side we get 2(b.c.Cos(A)+a.c.Cos(B)+a.b.Cos(C))=a^2+b^2+c^2
This will get to the result
Peter Tran
B, C, E, F are concyclic; AEC, AFB are secants
ReplyDeleteSo AB.AF = AC.AE. Similarly
BC.BD = BA.BF
CA.CE = CB.CD
2(AB.AF + BC.BD + AC.CE)
=(AB.AF + BC.BD + AC.CE)+(AB.AF + BC.BD + AC.CE)
=(AC.AE + BA.BF + CB.CD)+(AB.AF + BC.BD + AC.CE)
=(AC.AE + AC.CE)+(BF.BA+AB.AF)+(CD.CB+BC.BD)
=AC(AE+CE)+ AB(AF+BF)+BC(BD+CD)
=AC.AC + AB.AB + BC.BC
=a^2 + b^2 + c^2
=Sa + Sb + Sc