Saturday, February 20, 2010

Problem 427: Triangle, Two Altitudes, Square of a Side

Proposed Problem
Click the figure below to see the complete problem 427 about Triangle, Two Altitudes, Square of a Side.

Problem 427: Triangle, Two Altitudes, Square of a Side.
See also:
Complete Problem 427

Level: High School, SAT Prep, College geometry

8 comments:

  1. We've AD=bsin(C) & AE=bcos(A). Also AFcos(90-B)=bcos(A). So AF=bcos(A)/sin(B). Hence, AD*AF=b^2*sin(C)cos(A)/sin(B). Similarly, CF*CE=b^2*sin(A)cos(C)/sin(B) which gives us: AD*AF+CE*CF=(b^2/sin(B))[sin(C)cos(A)+sin(A)cos(C)]=(b^2/sin(B))[sin(C+A)]=b^2=AC^2 since A+B+C=180 deg.
    Ajit

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  2. ▲AFG ~ ▲BFD, ▲FGC ~ ▲EFB (right tr,& ang perpen sides) =>

    AF/BF = FG/FD => BF∙FG = AF∙FD (1)
    FC/BF = FG/EF => BF∙FG = FC∙EF (2)
    from (1) & (2)

    AF∙FD = FC∙EF

    AF∙FD + FC∙EF = 2FC∙EF
    AF( AD-AF ) + FC( EC - FC ) = 2FC∙EF
    AF∙AD - AF² + FC∙EC - FC² = 2FC∙EF
    AF∙AD + FC∙EC = AF² + 2FC∙EF + FC²
    AF∙AD + FC∙EC = AE² + EF² + 2FC∙EF + FC²
    AF∙AD + FC∙EC = AE² + ( EF + FC )²
    AF∙AD + FC∙EC = AE² + EC²

    AF∙AD + FC∙EC = AC²
    -----------------------------------------

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  3. trazas la perpendicular de F al lado AC y determinas dos cuadrilateros inscrisptibles y aplicas el teorema de las tangentes osea ;
    AF.AD=AH.AC CF.CE=CH.AC
    sumamos las dos ecuaciones :
    AF.AD + CF.CE = AC (AH + CH ) pero AH + CH = AC
    AF.AD + CF.CE = AC.AC --- Lqqd

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  4. era el teorema de las secantes plop -_-!! para la proxima reviso mejor antes de publicar

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  5. Nicely done in the Feb 21, 8:48 post above. I cannot read Spanish, but I think H is the foot of the perpendicular from F to AC. Then circumscribing circles about the cyclic quadrilaterals FEAH and FDHC leads to the first two product equations.

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  6. F is the orthocentre let BH meet AC at H.

    From similar triangles AF. AD = AH. AC and CF. CE = CH. AC

    Adding; AF. AD + CF. CE = AH. AC + CH. AC = AC ^2.

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. It follows that AF. AD + BF. BH + CF. CE = 2 (AB^2 + BC^2 + CA^2)

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  8. Is AD.AF=AB.AE , CE.CF=CD.CB ( E,B,D,F are conciclic )AB^2=AC^2+BC^2-2.CD.BC,
    BC^2=AB^2+AC^2-2.AE.AB so AC^2=CD.BC+AE.AB=CE.CF+AD.AF.

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