In ΔABC, point D lies on the side AB between vertices A and B. If CD =

From my reasoning, the correct answer is A.

Rephrasing the problem, we want to know if angle ACD > angle CDB because the opposite triangle sides of these angles are the unknowns in the question stem: AD > BC?

1). Knowing ADC = 110 means we know angle CDB is 70. Cool, so we know one of our 2 wanted values. What can we say about angle ACD? Well, since we know all three angles of a triangle must add up to 180 and one of the angles is already known to be 110, the other two angles can't add up to be greater than 70. Looking back at our question stem, let's try and prove that angle ACD CAN be greater than angle CDB. Thus ACD would have to be greater than 70. But, the max it can be is 69 since angle CAD can't be 0 degrees so the smallest it can be is 1 degree. This would answer the question stem with No. Any other valid value of angle ACD will be smaller than 69 and also answer No. Thus, this statement is sufficient to answer the question.

2). Knowing angle ACD doesn't tell us anything. We can't solve for any other angle or side of a triangle so this doesn't help bring us anywhere. Boo. Insufficient.

Answer is A.

Let me know if anyone else agrees / disagrees.