Circle Geometry with Cyclic Quadrilateral

[marthamatica]

Hi, I am stuck on this problem! I know that the opposite corners must add to 180. So the opposite corner from the 130 corner is 50 deg. Then because the angles are indicated to be equal there is another 50. The remaining angle of the triangle is 80. I cannot find the remaining angles inside the 130 deg corner triangle. I know they must sum to 50 because one triangle angle is 130.

HELP! I'm almost thinking this cannot be solved.

Pic attached.

 

[marthamatica]

Did this while watching the Ottawa : New Jersey hockey game...so no guarantees!!

Label the quadrilateral ABCD, B being the 130 degrees, A and D the 50 degrees.
Agree: angleACD = 80 degrees. Also, triangleACD is isosceles, since AC = CD.

Let AC = CD = 1.
Then AD = 2SIN(40) = ~1.2855 and circle radius = 1 / [2SIN(50)] = ~.6527

I'll leave you that to play with...

Thanks!!!

I agree with those calculations. I can't see how those help me to find the unknown angles. I have mostly been trying stuff with inscribed angles but nothing helps with the two unknown angles inside the ABC triangle.

 

[marthamatica]

Agree; don't think it's possible.
BUT it is easy enough to "see" that, if AB <> CB, then point B has 2 possible
locations; agree?

I don't know what '<>' means or why point B would have 2 possible locations.

I spent a while scratching my head over this problem. I think there should have been extra information given in the problem.

 

[marthamatica]

<> means not equal to.
Point B can be at 2 different locations on arcAC, left portion and right portion,
130 degrees both locations, by symmetry. OK?

ABC is an inscribed angle so B can move around the circumference and ABC will still have 130 deg in every location.