Friday, December 28, 2012

Problem 840: Parallelogram, Perpendicular, Diagonal, Metric Relations, Similarity

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 840.

Online Geometry Problem 840: Parallelogram, Perpendicular, Diagonal, Metric Relations, Similarity

6 comments:

  1. build DM and BH altitudes TO AC [CMD=DHB=90]
    DMC~AFC =>CM/FC=CD/AC
    AC*CM=FC*CD
    BHC~AEC=>BC/AC=HC/EC
    EC*BC=HC*AC
    AMD~=CHB =>CH=AM
    EC*BC+CD*FC=AC*CM+HC*AC=AC(CM+HC)=AC(CM+AM)=AC*(AC)=AC^2
    Q.E.D

    ReplyDelete
  2. By similar triangles ABE and ADF, we have
    BE/AB = DF/AD <-> BE/CD = DF/BC <-> BE*BC=CD*DF

    AC^2
    =AD^2 + CD^2 - 2AD*CD*cos∠ADC
    =BC^2 + CD^2 - 2BC*CD*cos∠ADC
    =BC^2 + CD^2 + 2BC*AB*cos∠ABE
    =BC^2 + CD^2 + 2BC*BE
    =BC^2 + CD^2 +BC*BE + CD*DF
    =BC*(BC+BE) + CD*(CD+DF)
    =BC*CE + CF*CD
    q.e.d.

    ReplyDelete
  3. Let G be a point such that EAGC is rectangle.
    Let M be a point on AC such that DM⊥AC.

    C,G,D,M concyclic ⇒ CE×BC = AD×AG = AM×AC
    A,F,D,M concyclic ⇒ CF×CD = CM×AC

    ∴ CE×BC + CF×CD = AC×(AM+CM) = AC^2

    ReplyDelete
  4. Confining to ΔABC, in the usual notation,
    b = a cos C + c cos A = a. CE/b + c. AF/b
    AC² = b² = a.CE + c. CF = CB. CE + CD. CF

    ReplyDelete
  5. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 30, 2012 at 2:22 AM

    BC^2=AF^2+FD^2=AF^2+(CF-DC)^2;DC^2=AE^2+EB^2=AE^2+(CE-BC)^2;
    AC^2=AE^2+EC^2;AC^2=AF^2+FC^2

    BC^2=AF^2+CF^2-2CF.DC+DC^2
    DC^2=AE^2+CE^2-2CE.BC+BC^2
    --------------------------------------------------
    2CF.DC+2CE.BC=(AF^2+FC^2)+(AE^2+EC^2)=2AC^2

    ReplyDelete
  6. Solution to problem 840 by Michael Tsourakakis for Greece
    ECFA,is ,cyclic quadrilateral.So EC.AF+AE.CF=AC.EF (1)More,angle AEF=angle ACF and
    angle AFE=angle ECA=angle CAD.So,The triangles AEF, ACD are similar.Therefore
    AC/EF=BC/AF=CD/AE.So,AE=CD.EF/AC and AF=BC.EF/AC
    from (1) we,CE.BC.EF/AC+DC.EF.CF/AC=AC ,EF(CE.BC+DC.CF)=EF.AC.AC or
    ,CE.BC+DC.CF=AC^2

    ReplyDelete