Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Ukraine

Let $AD$ be the median of a triangle $ABC$ , and let $\angle ADB = 45^\circ$ , $\angle ACB = 30^\circ$ . Find $\angle BAD$ (in degrees).

Solution

Нехай $BK$ — висота трикутника $ABC$ . Оскільки точка $D$ є серединою гіпотенузи $BC$ прямокутного трикутника $KBC$ , то трикутник $BKD$ є рівностороннім. Отже, як нескладно бачити, $\angle ADK = \angle DAK = 15^\circ$ , $AK = DK = BK$ . А тому $\angle BAK = 45^\circ$ , $\angle BAD = 45^\circ - 15^\circ = 30^\circ$ .

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