ACB is tangent to a circle at C. CD and CE are chords such that $\angle ACE > \angle ACD$. If $\angle ACD = \angle BCE = 50^{\circ}$, then which is correct answer?

By the alternate segment theorem, the angle between tangent AC and chord CD equals the angle in the alternate segment, so angle CED is 50 degrees. Since angle ACE is given as 50 degrees and angle CED is 50 degrees, the lines AB and ED are parallel. In triangle CDE, angle DCE equals 180 degrees minus angle ACE minus angle BCE, which equals 180 minus 50 minus 50 to give 80 degrees. Since angle CED is 50 degrees, angle CDE must be 50 degrees, meaning triangle CDE is isosceles with opposite sides CD and DE being equal.

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