Geometry Problem

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

Click the figure below to see the problem 723 details.

## Sunday, January 29, 2012

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

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First we observe that diagonal AOC is a diameter of circle (O) since <ADC is a rt angle.

ReplyDelete<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.

Solution 2:

ReplyDelete<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]

We can prove this problem using result of problem 722 or

ReplyDeleteNote 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

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

ReplyDeleteLet the interecting point of CG & EF be Y

ReplyDeleteLet <ECY=x

<EYC=90-x=<FYG

<CYF=180-<EYC=90+x

<CYF+<FYG=90+x+90-x=180

CHG is a st. line