Friday, August 29, 2008

Elearn Geometry Problem 166

See complete Problem 166 at:

Parallelogram, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. draw diagonal AC which meets BD at G and BE at H.also join GE.
    proof:we know that in a parallelogram diagonals bisect each AG=GC and BG=GD.
    also median of a triangle divides it into two triangles of equal area.
    [tri BGE]=[tri GED]
    [tri BGH]+[tri GHE]=[tri GEF]+[tri BFD]-----( 1 )
    [tri AEG]= [tri GEC]
    [tri AFG]+[tri GEF]=[tri GHE]+[tri HEC]------(2)
    On adding 1 and 2 equations
    [tri BGH]+[tri AFG]=[tri EFD]+[tri HEC]
    On adding [tri AGB] on either side
    [tri ABF]+[tri BGH]=[tri AGB]+[tri EFD]+[tri BHC]
    but diagonals of a parallelogram divide it into four triangles of equal area.
    [tri AGB]=[tri BGC] substitute in the above equation
    and after simplifications we get the required result. S=S1+S2

  2. Let A be (0,0), B;(b,h), C:(b+d,h) and D:(d,0) and let E be (p,q) Now we can determine the F as
    (dhp/(ph-qb+qd),dhq/(ph-qb+qd)). Now Tr. AFD =d*dhq/2(ph-qb+qd)= hqd^2/(ph-qb+qd) -----(1)
    While, Tr. BFE =(1/2)(bdhq/(ph-qb+qd) -bq+dhpq/(ph-qb+qd)-pdh^2/(ph-qb+qd) +ph -pdhq/(ph-qb+qd))and Tr. ECD =(dh+bq-hp)/2
    Now Tr. BFE + Tr. ECD = hqd^2/(ph-qb+qd) upon addition and simplification. In other words, Tr. AFD = Tr. BFE + Tr. ECD because of (1)
    In parallelogram ABCD, Tr. ABD = Tr. BDC or S + Tr. AFD = S1 + S2 + Tr. BFE + Tr. ECD or S = S1 + S2 since Tr. AFD = Tr. BFE + Tr. ECD

  3. Es mucho más fácil: aplicación directa del teorema de la alfombra.
    Por un lado: [ABD]= S+[AFD]=0'5*[ABCD]
    Y por otro: S1+S2+[AFD]=0'5*[ABCD]
    Por tanto, S=S1+S2

  4. Solution to problem 166.
    Let be h1 and h2 the distances from E to lines BC and AD, respectively, and h3 the distance from F to line AD. We have BC = AD.
    Then S1 = BC.h1/2 = AD.h1/2,
    S2 = S(ADE) – S(ADF) = AD.h2/2 – AD.h3/2
    and S = S(ABD) – S(ADF) = AD(h1+h2)/2 – AD.h3/2 = AD(h1+h2-h3)/2.
    Hence S1 + S2 = AD.h1/2 + AD.h2/2 – AD.h3/2 = AD(h1+h2-h3)/2 = S.

  5. To Antonio Ledesma:
    Why is it S1+S2+[AFD] = 0,5*[ABCD]?
    (I know that "alfombra" means carpet - "tapete", in portuguese, but I don't know the "alfombra" theorem)