Geometry Problem

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## Saturday, February 26, 2011

### Problem 584: Cyclic Quadrilateral, Diagonals, Circumcenters, Circumcircles, Parallelograms

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EFGH from 583

ReplyDeleteName E', H', G' midpoint of AB, AD, CD

AEM = 2ABM, DGM = 2MCD => AEM = DGM (1)

From AEOH', DGOH' AEO = DGO (2)

From (1) & (2) MEO = MGO (3)

From AE'OH', DG'OH' EOG = A+D (4)

▲BEM => BME = 90 - BAM, ( BEM = 2BAM )

▲CMG => CMG = 90 - CDM

▲BMC => AMD = 180 - MAD - MDA

=> EMG = A+D (5)

From (4) & (5) EOG = EMG (6)

From (3) & (6) => EOGM parallelogram

Problem 584 solution: This solution was submitted by Michael Tsourakakis from Greece

ReplyDeleteThanks Mixalis.