2014 AMC 8 Problem 9

Attempt Problem 9 of the 2014 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

9.

In △ABC,\triangle ABC, DD is a point on side AC‾\overline{AC} such that BD=DCBD=DC and ∠BCD\angle BCD measures 70∘.70^\circ. What is the degree measure of ∠ADB?\angle ADB?
100 100
120 120
135 135
140 140
150 150
Answer: D
Concepts:isosceles triangleangle chasing
Small hint:
Since BD=DCBD=DC, triangle BCDBCD is isosceles.
Big hint:
Find ∠BDC\angle BDC, then use the straight angle at DD.
Solution:
Since BD=DCBD=DC, triangle BDCBDC is isosceles, so ∠DBC=∠BCD=70∘\angle DBC=\angle BCD=70^\circ.
Thus ∠BDC=180∘−70∘−70∘=40∘.\begin{aligned}\angle BDC&=180^\circ-70^\circ-70^\circ\\&=40^\circ.\end{aligned} Because A,D,CA,D,C are collinear, ∠ADB\angle ADB and ∠BDC\angle BDC form a straight angle, so ∠ADB=180∘−40∘=140∘\angle ADB=180^\circ-40^\circ=140^\circ.
Thus, D is the correct answer.
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