I have trouble finding angle BCE. This question right here is very confusing, and I have been trying so hard, but to no avail. Please help me.
Chris told me this by message.
I believe that quadrilateral is inscribed in a circle....I think CE is the minor arc that the angle subtends....thus the angle itself = 25 degrees. I know the angle at the"bottom" of the quad has to be greater than 65 degrees and that the point "D""?? is the midpoint on that chord. After that, I'm stuck!!
You seem to me making an awful lot of assumptions there Chris!
I'm pretty sure arc m CE=50 means the arc measure of CE = 50
Also, I'm not sure if it helps but <A and <C have to be the same because it's an isoceles triangle.
Like I said before, I think that arc m CE=50 means the arc measure of CE. If a circle is drawn within the the quadrilateral, the arc of CE if 50. I also went ahead and drew an axillary line from angle B to angle E that passes through point D. If arc CE = 50 than angle CDE should equal 50. Vertical angles are congruent, which makes angle BDA 50 which makes its arc BA 50. Angle EDA equals 130 because it is supplementary to Angle CDE which is 50. Since Angle EDA is 130 then its arc AE 130. If you combine arc AE and arc BA then arc BAE equals 180. Angle BCE is 180 degrees because its arc is 180 degrees.
There is a flaw in my final answer, the angle measure would have to be half the measure of the arc because it's no longer an inscribed angle. Angle BCE is 90 degrees.