Friday, May 23, 2025

Geometry Problem 1601: Rectangular Property of Cyclic Orthodiagonal Quadrilaterals

Challenging Geometry Puzzle: Problem 1601. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.

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Illustration of Geometry Problem 1601

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1 comment:

  1. < ABP = < ACD = < DPJ = < BPK
    Hence KB = KP and so K is the midpoint of AB

    Similarly for M,F,H

    Hence KM//AC//HF and KH // BD//MF and so
    KMFH is a parallelogram

    Since AC is perpendicular to BD it follows that
    KMFH is a Rectangle

    Sumith Peiris
    Moratuwa
    Sri Lanka

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