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 1226.
Friday, June 10, 2016
Geometry Problem 1226: Quadrilateral, Squares, Centers, Midpoints, Congruence
Subscribe to:
Post Comments (Atom)
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.