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Geometry ProblemProblem submitted by Jacob Ha, Keith Law, and more.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 831.
Solution of myself: Rotate 180° about O, then D→B, B→D. Let C→C', then C' is a point on line OA, with OC'=OC, so BCDC' is a parallelogram. Since ∠BC'D=∠BCD=∠BAD, C' coincide with A. Therefore, ABCD is a parallelogram. ******Solution submitted by Bean Ng: If OA=OC, then we are done. Otherwise, WLOG, suppose OA>OC, choose a point X on OA such that OX=OC. Then XBCD is a parallelogram. So ∠BXD=∠BCD=∠BAD. On the other hand, we have ∠BXD=∠BAD+∠ABX+∠ADX, thus ∠ABX=∠ADX=0. Hence, A=X (two points coincide)and so ABCD is parallelogram.
I hope others will give some more elementary solution. i.e. not involve uniqueness argument.
vectorAD+vectorDC=vectorACvectorAB+vectorBC=vectorACtherefore, AD+DC=AB+BCvectorBA+vectorAD=vectorBDvectorBC+vectorCD=vectorBDtherefore, BA+AD=BC+CDsince ∠BAD=∠BCDthus, AD=BC and AB=DCand so ABCD is a parallelogram.
Actually this problem is come from a variation of Problem 782 & 783. http://gogeometry.com/school-college/p782-triangle-orthocenter-circumcircle-diameter-parallelogram.htmhttp://gogeometry.com/school-college/p783-triangle-orthocenter-circumcircle-diameter-midpoint.htm***The original problem is, as shown in 783: Let M be the mid-point of BC, produce HM to D, such that HM=MD, then BDCH is a parallelogram. Thus, ∠BDC=∠BHC. But ∠BHC=180°−∠A, so ∠BDC=180°−∠A, therefore ABDC is concyclic. Consequently, D lies on the circumcircle. ***Now is my variation: Let M be the mid-point of BC, produce HM to meet the circumcircle at D. I want to show that BDCH is a parallelogram. And it is somehow a converse of the above problem. Now I have BM=MC, and ∠BDC=180°−∠A=∠BHC, and hence Problem 831. I hope this can help anyone to give a elementary solution.