## Thursday, December 13, 2012

### Problem 831: Quadrilateral, Diagonal, Midpoint, Opposite Angles, Congruence, Parallelogram

Geometry Problem
Problem 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.

1. 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.

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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.

On the other hand, we have

Hence, A=X (two points coincide)
and so ABCD is parallelogram.

2. I hope others will give some more elementary solution.
i.e. not involve uniqueness argument.

vectorAB+vectorBC=vectorAC
vectorBC+vectorCD=vectorBD
and so ABCD is a parallelogram.

4. Actually this problem is come from a variation of Problem 782 & 783.

http://gogeometry.com/school-college/p782-triangle-orthocenter-circumcircle-diameter-parallelogram.htm

http://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.