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

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## Thursday, December 2, 2010

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Problem 545: Acute Triangle, Squares, Altitudes, Area

Online Geometry theorems, problems, solutions, and related topics.

Geometry Problem

Click the figure below to see the complete problem 545 about Acute Triangle, Squares, Altitudes, Area.

Go to Complete Problem 545

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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