Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, June 10, 2016

### Geometry Problem 1226: Quadrilateral, Squares, Centers, Midpoints, Congruence

Labels:
center,
congruence,
midpoint,
quadrilateral,
square

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Per van Aubel's theorem, Lines O1O3 and O2O4 are equal length and cross at right angle

ReplyDeleteWe have M1M2=1/2. O1O3=M3M4 and M1M2//O1O2//M3M4

and M1M4= 1/2. O2O4=M2M3 and M1M4//O2O4//M2M3

so M1M2M3M4 is a square

That M1M2M3M4 is a parallelogram is easily seen from the midpoint theorem

ReplyDeleteTo show that this parallelogram is indeed a square we need to show that O1O3 and O2O4 are equal and cut at right angles - this follows easily from Van Aubel's theorem.