See complete Problem 33 at:

www.gogeometry.com/problem/p033_triangle_quadrilateral.htm

Triangle and Quadrilateral angles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 33: Triangle and Quadrilateral

Labels:
angle,
quadrilateral,
triangle

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I have this question:

ReplyDeleteIf K is the circumcircle of BCD, why isnt direct that A is the center? I mean: angles BAC and BDC have same arc and one is half of the other...

Consider the circumcircle of triangle BAD, then, arc BD is the locus of all P points such that <CPB=2<CDB

DeleteGreetings :)

Let B' is reflection of B in AD.

ReplyDeleteThen, Angle AB'D=Angle DCE, so B,C,D,B' is cyclic.

Also, Angle BDC=Angle BB'C=theta=Angle B'CA (because Angle BCD=2*theta) It means that AB'=AC. and BA=B'A by B' is reflection of B.

So, A is center of circle passing B,C,D,B'. Therefore AD=AB, x=45 degree.

Extend BA to F such that BA = AF. Then < DFA = x and hence BCDF is con cyclic. So < BFC = theta, hence < ACF = theta. So BA= CA= FA and hence < BCF = 90

ReplyDeleteNow < FCD = < FBD = < DCE = x, hence 2x= 90 and x=45

Sumith Peiris

Moratuwa

Sri Lanka

This is true irrespective of the value of theta.

ReplyDelete