Wednesday, March 23, 2016

Geometry Problem 1202: Four Squares, Two Equal Parallelograms, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1202.

Geometry Problem 1201: Triangle, Interior Perpendicular Bisector, Midpoint, 90 Degrees, Area, Online College School


  1. Dear Antonio - By "equal" I presume u mean equal in area and not congruent.

    Let the square of diagonal AB be of side a, BC be of side b, CD be of side c and DA be of side d.

    Triangles AA1D2 and AD1A2 are congruent SAS so < A1D2A = < AD1A2 = e say....(1)

    Similarly triangles DD2C1 and DD1C2 are congruent SAS (since < C2DD1 = < C1DD2) and so < DD2C1 = < DD1C2 = g say ....(2)

    From (1) and (2), < C2D1A2 = e+g-90....(3) and
    < C1D2A1 = 270-e-g...(4)

    Now the angles in (3) and (4) are supplementary so S(C2D1A2) = S(C1D2A1) ...(5) using area = 1/2 ab sin@ and sin@ = sin (180-@)

    Similarly we can show that in triangles A2B1C2 and A1B2C1 that B1C2 = B2C1, B2A1 = B1A2 and the included angles < C2B1A2 and < C1B2A1 are supplementary so that

    S(A2B1C2) = S(A1B2C1) ...(6)

    (5)+(6) => S(A2B1C2D1) = S(A1B2C1D2) i.e. the 2 quadrilaterals are equal.

    The 4 sides of 1 quadrilateral are the same as the 4 sides of the other
    Each of the corresponding included angles are supplementary

    Yet the fact that these are parallelograms I'm yet to prove.

    Sumith Peiris
    Sri Lanka

  2. Dear Sumith
    Thanks for your comment. I have changed the notation to
    "congruent (same shape and size) parallelograms"


    1. triangle BB1A2 congruent to BB2A1 ( case SAS) => B1A2=B2A1=u
    Similarly B2C1=B1C2= v and A1D2=A2D1= x and D1C2=D2C1= w
    2. Triangle CB2C1 similar to CBD ( case SAS) with similar factor = sqrt(2) => B2C1=v= BD/sqrt(2)
    Triangle AA1D2 similar to ABD ( case SAS) with similar factor = sqrt(2) => A1D2=x= BD/sqrt(2)=B2C1
    Similarly we have B1A2=A1B2=u=AC/Sqrt(2)= D1C2=C1D2
    Quadrilateral A1B2C1D2 have opposite sides congruent so A1B2C1D2 is a parallelogram
    Similarly Quadrilateral C2B1A2D1 is a parallelogram
    3. Consider quadri. ABCD with diagonals BD and AC .
    Define angles α, β, γ, θ as shown on the sketch.
    Angle (A1B2C1)= 360- 90- (α+ β)= 270- (α+ β)
    Angle B1C2D1)= 90+( γ+ θ)
    Note that α+ β+ γ+ θ= 180 so Angle B1C2D1= 90+ 180-(α+ β)= angle (A1B2C1)
    Parallelograms A1B2C1D2 and C2B1A2D1 have all coresponding sides and angles congruents so these parallelograms are congruent