See complete Problem 35 at:

www.gogeometry.com/problem/p035_incenter_cyclic_quadrilateral.htm

Incenters and Inradii in Cyclic Quadrilateral. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 35: Quadrilateral, Circle

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(1) Since E is an incenter of triangle BD,

ReplyDeleteang(AED)=90°+(ang(ABD)/2)=90°+(ang(ACD)/2),

ang(HED)=ang(HAD)=ang(CAD)/2,

ang(BEA)=90°+(ang(BDA)/2)=90°+(ang(BCA)/2) and

ang(FEB)=ang(FAB)=ang(CAB)/2.

Hence we see that

ang(FEH)

=360°-(ang(AED)+ang(HED)+ang(BEA)+ang(FEB)

=360°-

(180°+(ang(ABD)+ang(CAD)+ang(BDA)+ang(CAB))/2

=360°-(180°+(ang(ABC)+ang(ADC))/2

=360°-(180°+90°)

=90°

and hence EFGH is a rectangle.

(2) We get

(i) By (1), EH^2 =FG^2.

(ii) EH^2

=(r_d -r_a)^2 +(AD-(AB+AD-BD)/2-(DA+DC-AC)/2 )^2

(iii) FG^2

=(r_c -r_b)^2 +(BC-(BA+BC-AC)/2-(CB+CD-BD)/2 )^2

Hence we have (r_d -r_a)^2 =(r_c -r_b)^2,

that is, r_a +r_c =r_b +r_d

I'd like to know why is it ang(HED)=ang(HAD), in the solution presented by bae deok rak. I thank you in advance.

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