PyjamaScientist
In \(\triangle \)ABC point D is on the side AB between vertices A and B. If CD = AB, is AD > BC?
(1) \(\angle\)ADC = 110\(^{\circ}\)
(2) \(\angle\)ACD = 30\(^{\circ}\)
Statement 1 Alone:First, draw AB. Pick any point D on AB and draw CD = AB with a 110-degree angle in any direction. Complete the rest of the triangle by connecting AC and BC.
Then we just proved we can pick any point D on AB, therefore statement 1 alone is insufficient.
Statement 2 Alone:Draw AB, then pick any point D on AB and draw CD = AB. Connect AC, then rotate CD with respect to point D so that \(\angle\)ACD = 30\(^{\circ}\). Finally, connect BC.
Then we just proved we can pick any point D on AB, therefore statement 2 alone is insufficient.
Both Statements Combined:Draw triangle ACD first with the 30 and 110-degree angles. Extend AD to AB so that AB = DC and connect BC.
Then we just proved the shape has all fixed angles and we cannot change the ratios of lengths. Thus combined it is sufficient.
Answer: C