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If AB = AC and ∠1 > ∠B, then

∠1 > ∠B
Since AB = AC
∠C = ∠B

∠1 > ∠C

IMO B
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Bunuel

If AB = AC and ∠1 > ∠B, then

(A) ∠B > ∠C
(B) ∠1 > ∠C
(C) BD > AD
(D) AB > AD
(E) ∠ADC > ∠ADB

Attachment:
2020-12-10_13-10-12.png

The only option which is correct in my opinion is D

(A) ∠B > ∠C, since, AB = AC so, ∠B = ∠C (Incorrect)
(B) ∠1 > ∠C. Since we are not sure whether the angle is acute or obtuse, and, BD > DC or, DC > BD (No evidence)
(C) BD > AD (No evidence)
(D) AB > AD, is true, 1. Perpendicular distance is the shortest distance (based on that logic), or 2. 180 - ∠1 > ∠B
(E) ∠ADC > ∠ADB (same as option B, No evidence)
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Isn't the best way here to visualize? If ∠1 > ∠B then that means AD will shift closer to C (making it increasingly large/obtuse).

As a result ∠1 > ∠C.

B.

Over simplifying?
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