Friday, July 7, 2017

Geometry Problem 1338: Four Squares, Diagonals, Angle, 45 Degrees, Areas

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

Details: Click on the figure below.

Geometry Problem 1338: Four Squares, Diagonals, Angle, 45 Degrees, Areas.



    Let u= ∠ (FAD) and v= ∠ (BAE)
    Create triangle ADP such that triangle ABE congruent to tri. ADP( see sketch)
    We have ∠ (DAP)= v and ∠ (ADP)= 45 and AE=AP, BE= DP
    U+v= 45 and FD⊥DP
    Triangle AEF congruent to tri. APF ( case SAS)
    So EF= FP
    Applying Pythagoras theorem in triangle FDP
    FP^2= FD^2+DP^2 = EF^2= FD^2+BE^2
    From this result we will have S= S1+S2

  2. Problem 1338
    Let AE ,BC intersects in K (K on BC) and AF,CD intersects in L ( L on CD).
    Let AM ( M on extention CD) with AM perpendicular in AE.
    Now triangle ABK=triangle ADM.Let BN perpendicular in FD (N on AM).
    But triangle ABE=triangle and AE=AN.
    Is triangle AEF=triangle AFN. So EF=FN, <FDN =90.
    FN^2=FD^2+DN^2 or EF^2/2=FD^2/2+BE^2/2 or S=S1+S2.

  3. Find G such that < GAE = < GDF = 90
    Let BE = a, EF = b and FD = c

    < FAG = 45 and < BAE = < DAG so Tr.s ABE & ADG are congruent ASA

    So AG = AE
    Hence Tr.s AEF & AGF are congruent SAS and hence FG = b

    Applying Pythagoras to Tr. DFG,
    a^2+c^2 = b^2

    But S1 = a^2/4 and similar for b^2 and c^2
    So S1+S2 =S

    Sumith Peiris
    Sri Lanka

  4. Let the lower left point of the yellow square be M.
    By use of the inscribe angle theorem we see that M is the center of the circumscribed circle of triangle AEF. As a result AM = EM.
    Let N be the intersection of the AD and the line through EM.
    Now apply Pythagoras on triangle ANM

    1. Top Class Deduction....!!

  5. We shall prove BE^2+DF^2=EF^2 (*), equivalent to relation to be proven.
    Let G reflection of B in AE; easily we notice it is also the reflection of D in AF, thus by symmetry <AGE=<ABE=45 and GE=BE, <AGF=<ADF=45 and FG=FD hence
    <EGF=90. By Pythagorean theorem in triangle EGF we get (*), thus we are done.

    Best regards

  6. spin the figure 90 degrees clockwise around A.
    Let E' be the point correponding to the rotation of E.
    E'D = EB cause D is rotation of B.
    triangle FE'D is rectangle in D.
    Triangle EAF =Triangle AFE' AF is common, AE=AE' and both have 45 degree between these sides
    Apply Pythagorean theorem to FE'D

  7. Let the lower left points of S1 and S2 be K and L respectively, the top right point of S be T, and AH be the altitude from A, i.e. AH _|_ BD and H is on BD. Since AK/AH = AE/AF = AH/AL => AH^2 = AK.AL (1). Since AB = AD, the area of triangle ABD = AH.BD/2 = AH^2 (2). Using (1) and (2) we have that area of AKTL = AK.AL = AH^2 = area of triangle ABD = area of AKEFL + area of triangle KBD + area of triangle LFD = (area of AKTL - S/2) + S1/2 + S2/2 => 0 = -S/2 + S1/2 + S2/2 => S = S1 + S2.