## Sunday, January 29, 2012

### Problem 723: Squares, Circumscribed Circles, Collinearity

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

Click the figure below to see the problem 723 details.

1. First we observe that diagonal AOC is a diameter of circle (O) since <ADC is a rt angle.
<AHC = <ADC = 90 deg
Similarly <EHG = < EFG = 90 deg
By Problem 722,
A, E, H are collinear and so <AHG is the same as <EHG = 90 deg
Hence <AHC + <AHG = 180 deg and
C, H, G are collinear.

2. Solution 2:
<DHG = <DEG = 45 deg
<DHC = supplement of <DBC = 180deg - 45deg
Sum = 180 deg
So G, H, C are collinear.
[Note: DB passes thro'O and bisects the rt <ABC]

3. We can prove this problem using result of problem 722 or
Note that DF and DB are diameters of circles O and O’
So (BHD)=(DHF)=90 >> B,H, F are collinear
(GHF)=(CHB)=45 ----- (angles face 90 arc)
So (CHG)=(BHF)+45-45= 180 >> C, H, G are collinear

4. Reference 722 < AHC = 90 = < EHG. Hence CHG are collinear

5. Let the interecting point of CG & EF be Y
Let <ECY=x
<EYC=90-x=<FYG
<CYF=180-<EYC=90+x
<CYF+<FYG=90+x+90-x=180
CHG is a st. line