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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.
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 thatS(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 otherEach of the corresponding included angles are supplementary Yet the fact that these are parallelograms I'm yet to prove.Sumith PeirisMoratuwaSri Lanka
Dear Sumith Thanks for your comment. I have changed the notation to "congruent (same shape and size) parallelograms"
http://s30.postimg.org/5zl4in7i9/pro_1202.png1. triangle BB1A2 congruent to BB2A1 ( case SAS) => B1A2=B2A1=uSimilarly B2C1=B1C2= v and A1D2=A2D1= x and D1C2=D2C1= w2. 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)=B2C1Similarly we have B1A2=A1B2=u=AC/Sqrt(2)= D1C2=C1D2Quadrilateral A1B2C1D2 have opposite sides congruent so A1B2C1D2 is a parallelogramSimilarly Quadrilateral C2B1A2D1 is a parallelogram3. 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