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30 Jun 2022, 10:00
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Difficulty:
35%
(medium)
Question Stats:
68% (01:39) correct
32%
(02:13)
wrong
based on 22
sessions
History
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Not Attempted Yet
In triangle ABC, BC is extended to D. If ∠A = 50° and ∠ACD = 120°, then
(A) BC > AB
(B) AC > AB
(C) BC > AC
(D) AB > AC
(E) ∠B < ∠A
(A) BC > AB
(B) AC > AB
(C) BC > AC
(D) AB > AC
(E) ∠B < ∠A
30 Jun 2022, 10:34
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Bunuel
Alright, I'm going to be a pit of a pain in the butt. Only because I think there's at least as much to be learned about being thoughtful about possible rearrangements/manipulations on geometry questions than there is to be learned from the actual math in this specific question.
When BC is extended to D, it seems from the question that the intention is that they go in order BCD, but I'm not sure it's disallowed to go CBD.
Figure 1 attached goes with the obviously-intended case, and from that drawing, we know that AC>AB since ∠B is greater than ∠A. So, answer choice B.
But if we look at Figure 2, answer choice D would work.
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