Friday, September 5, 2008

Elearn Geometry Problem 171

Trapezoid, Midpoints, Triangles, Areas

See complete Problem 171 at:

Trapezoid, Midpoints, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. by Proposed Problem 170,

  2. See the drawing : Drawing

    - Define S0=S(ABCD), L midpoint of CD and h the height of CBE and DAE
    - S(ECL)=EL*h/2 and S(ELD)=EL*h/2 => S(ECD)=EL*h
    - S(BCE)=BC*h/2 and S(AED)=AD*h/2
    - EL=(BC+AD)/2
    - S=EL*h + (BC+AD)*h/2 => (BC+AD)*h => S(ECD)=S0/2
    - S(AIG)=S(GID) and S(BIF)=S(FIC) => S(ABFG)=S(DCFG) => S(DCFG)=S0/2
    - S(DCFG)= S(ECD) =>S1+S(CHMD)+S2= S(CHMD)+S
    - Therefore S=S1+S2

  3. See the drawing

    Define S0=[ABCD], L midpoint of CD, I intersection of AB and CD and h the height of CBE and DAE
    [ECL]=EL*h/2 and [ELD]=EL*h/2 => [ECD]=EL*h
    [BCE]=BC*h/2 and [AED]=AD*h/2
    S0=[ECD]+[BCE]+[AED]=EL*h + (BC+AD)*h/2= EL*h + EL*h
    => [ECD]=[BCE]+[AED]=S0/2
    [AIG]=[GID] and [BIF]=[FIC] => [ABFG]=[DCFG]
    S0=[ABFG]+[DCFG] => [DCFG]=S0/2
    S0/2=[DCFG]= [ECD] =>S1+[CHMD]+S2= [CHMD]+S
    Therefore S=S1+S2