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.

## Thursday, December 13, 2012

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

Labels:
angle,
congruence,
diagonal,
midpoint,
opposite,
parallelogram,
quadrilateral

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Solution of myself:

ReplyDeleteRotate 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.

ReplyDeletei.e. not involve uniqueness argument.

vectorAD+vectorDC=vectorAC

ReplyDeletevectorAB+vectorBC=vectorAC

therefore, AD+DC=AB+BC

vectorBA+vectorAD=vectorBD

vectorBC+vectorCD=vectorBD

therefore, BA+AD=BC+CD

since ∠BAD=∠BCD

thus, AD=BC and AB=DC

and so ABCD is a parallelogram.

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

ReplyDeletehttp://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.