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

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

Click the figure below to enlarge the problem 938.

## Wednesday, December 4, 2013

### Geometry Problem 938: Cyclic Quadrilateral, Triangle, Circumcircle, Circles, Circumcenter, Circumradius, Distance, Metric Relations

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Conjecture: O₁OO₂E form a parallelogram.

ReplyDeleteIf so, then using parallelogram law,

EO² + 9² = 2(4² + 6²)

EO = √23

Unfortunately, I still need some time to prove / disprove my assertion.

Let <BAE=a and <EBA=b. Then <OO1E=1/2<BO1A-<BO1E=(a+b)-2a=b-a. From cyclic condition we find that <EO2O=1/2<CO2D-<CO2E=b-a as well. If AB and CD meet at point P, <DPB=<DCA-<BAE=b-a. <O1OO2 is supplementary to <DPB because OO1 and OO2 perpendicular to AB and CD respective. That makes EO2OO1 a parallelogram.

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