Sunday, April 9, 2017

Geometry Problem 1327: Two Squares Side by Side, Perpendicular, 90 Degrees, Angle

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 1327: Two Squares Side by Side, Perpendicular, 90 Degrees, Angle.

Posted by Antonio Gutierrez at 10:25 AM
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5 comments:

  1. APOSTOLIS MANOLOUDISApril 9, 2017 at 11:46 AM

    Problem 1327
    Is triangle AED=triangle CED then AE=CG and <EAD=<DCG. But <DCG+<DGC=90.
    So <DAE+<DGC=90.Therefore AE is perpendicular at CG.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  2. APOSTOLIS MANOLOUDISApril 9, 2017 at 12:02 PM

    Problem 1327
    Is triangle AED=triangle CED(AD=DC,DE=DG, <ADE=90=<CDG) then AE=CG and <EAD=<DCG. But <DCG+<DGC=90.
    So <DAE+<DGC=90.Therefore AE is perpendicular at CG.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  3. c.t.e.oApril 9, 2017 at 12:23 PM

    Draw circles through ABCD and EFGD. L' intersection point.
    ang DL'G=45°, ang DL'C=135°=> L≡L' => ALC=90°

    ReplyDelete
  4. Sumith PeirisApril 9, 2017 at 1:32 PM

    Tr.s ADE & CDG are congruent SAS

    Hence < DAE = < DCG
    So ADLC is concyclic

    Therefore < ALC = < ADC = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Stan FulgerApril 9, 2017 at 2:49 PM

    Rotate triangle CDG by 90 degs about D so that G goes to E, C to A, done.

    ReplyDelete

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