Saturday, October 29, 2016

Geometry Problem 1280 Quadrilateral, Perpendicular Diagonals, Isosceles Right Triangles, 45 Degrees, Collinear Points

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 1280.

Geometry Problem 1280: Quadrilateral, Perpendicular Diagonals, Isosceles Right Triangles, 45 Degrees, Collinear Points.

4 comments:

  1. https://goo.gl/photos/3q9h6djv9nLppAob9

    Let M and N are the midpoint of BC and AD
    We have MG=MB=MC=ME
    And NF=NA=ND=NE
    So M and N are the centers of quadrilateral BEGC and AEFD
    In qua. BEGC we have ∠CEG=∠CBG= 45 degrees
    In qua. AEFD we have ∠FED=∠EAD= 45 degrees
    So F and G are located on angle bisector of angle CED => E, F and G are colinnear

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  2. Problem 1280
    Is <AED=AFD=90 so AEFD is cyclic,then <CEF=<FDA=45. But <BEC=<BGC=90 so BEGC is cyclic, then <GEC=<GBC=45=<CEF .Therefore the points E,F and G are collinear.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  3. Suppose that F is not on EG and that F' is the point at which AD subtends 90 on line EG. EF'G are collinear.
    Let X be any point on FE extended.

    Since BCGE is concyclic < XEB = 45
    Hence < XEA = 45 = < F'DA since AEF'D is concyclic.

    So F and F' must coincide and E,F,G are hence collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. ∵ The diagonals AC and BD are perpendicular at E
    ∴ ∠BEC = 90°
    ∵ AFD and BGC are isosceles right triangle
    ∴ ∠BGC = 90°
    ∴ ∠BEC = ∠BGC
    ∴ B, E, G and C are concyclic
    ∴ ∠GEC = ∠GBC
    ∵ AFD and BGC are isosceles right triangle
    ∴ ∠GBC = ∠GCB
    ∠GBC + ∠GCB + ∠BGC = 180°
    2∠GBC + 90° = 180°
    ∠GBC = 45°
    ∴ ∠GEC = 45°
    Similarly, ∠DEF = 45°
    ∵ The diagonals AC and BD are perpendicular at E
    ∴ ∠CED = 90°
    ∠DEF +∠CEF = 90°
    45° +∠CEF = 90°
    ∠CEF = 45°
    ∵ ∠CEG = ∠CEF
    ∴ The points E, F, and G are collinear

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