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.

6 comments:

  1. https://goo.gl/photos/dB3G7J8HWSRkdPrx9

    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

    ReplyDelete
  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.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  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
    Moratuwa
    Sri Lanka

    ReplyDelete
  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

    ReplyDelete
  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

    ReplyDelete