Thursday, April 23, 2020

Geometry Problem 1472: Cyclic Quadrilateral, Perpendicular Diagonals, Rectangle

Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

Dynamic Geometry Problem 1472: Cyclic Quadrilateral, Perpendicular Diagonals, Rectangle, Step-by-step Illustration, Step-by-step Illustration, iPad.

Posted by Antonio Gutierrez at 12:20 AM
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4 comments:

  1. c.t.e.o.April 23, 2020 at 5:18 AM

    ∆APE1, ∆PBE1 isosceles => E1 midpoint, ∆PCE2, ∆PBE2 isosceles => E2 midpoint
    => E1E2 midlle line of ∆ABC , at the same three rest sides
    https://photos.app.goo.gl/9X2XxujZ82wfkgDy9

    ReplyDelete
  2. c.t.e.o.April 23, 2020 at 7:53 AM

    https://photos.app.goo.gl/UCbnTTv6vi2aEGSK6

    ReplyDelete
  3. Sumith PeirisApril 23, 2020 at 12:28 PM

    < E1BP = < PCD = < DPH3 = < BPE1 so BE1 = PE1 and so E1 is the midpoint of AB
    Similarly E2, E3 and E4

    Hence E1E2 // AC // E3E4 & similarly E1E4//BD//E2E3

    Therefore E1E2E3E4 is a parallelogram & since AC//BD, it follows that E1E2E3E4 is a rectangle

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
    Replies
    1. Sumith PeirisApril 25, 2020 at 11:57 PM

      AC perpendicular to BD and not AC//BD in the last line of my proof above

      Delete

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