"Simple" geometry problem

coolul007

It doesn't equal 2r, my comment was if it does then that's the answer. What makes this problem "impossible" is that H or J is not fixed. phi becomes some angle but we don't know where the vertex is. AD can be thought of as a radius going through an angle of zero with AC through theta/2 to AI. Without H or J fixed at some distance this can't be solved. We were given a two stipulation fact: JD bisects delta AND meets AE at angle phi.

nburo

F, H and J are fixed. I'm studying this particular case.

Studiot

You have 10 (relevant) variables in a quadrilateral.

4 side lengths
4 vertex angles
2 diagonal lengths

You say you know four of these -one angle, two sides and one diagonal.

Attached are six equations (10 including the knowns) for the 6 unknowns.

These equations apply to a general quadrilateral. They cna be greatly simplified if the vertices can be shown to lie on a circle as I have already mentioned.

I have relabelled them to conform to conventional order. Theta is conventionally used for the intersection angle of the diagonals.

You can solve them at your leisure.

Attached Thumbnails

 

nburo

Thank you, but I think one of your 6 equations might be dependent of the others. (equation 6, I think)