Geometry Problem

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

Click the figure below to see the complete problem 907.

## Sunday, August 4, 2013

### Problem 907: Bicentric Quadrilateral, Incircle, Circumcircle, Circunscribed, Inscribed, Tangent, Inradius

Labels:
bicentric quadrilateral,
circumcircle,
incircle,
inradius,
tangent

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build CI IG and AI

ReplyDeleteBAH=2a=>C=180-2a

IAH=a, ICG=90-a

CGI=IHA=90=>CIG=a=IAH

=>CIG~IAH

1/x=x/3=>x=sqrt(3)

Minor correction due to typo error.

ReplyDeleteIn right triangle IHA we have IH=x= 3*tan(A/2)

in right triangle IGC we have x= CG*tan(C/2)= 1/tan(A/2) ...(A supplement to C)

Replace it in 1st expression we have x=3(1/x) => x=SQRT(3)

Peter Tran