Questions: If the angle ABD is 60 degrees, what is angle BAC? Show all your work.

If the angle ABD is 60 degrees, what is angle BAC? Show all your work.
Transcript text: If the $\angle A B D$ is 60 degrees, what is $\angle B A C$ ? Show all your work.
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If ABD\angle ABD is 60 degrees, what is BAC\angle BAC? Given ABD=60\angle ABD = 60^{\circ} Since B, D, and C are collinear, ABC=180\angle ABC = 180^{\circ}. Thus, DBC=180ABD=18060=120\angle DBC = 180^{\circ} - \angle ABD = 180^{\circ} - 60^{\circ} = 120^{\circ} In ABC\triangle ABC, AD is the altitude, and D is the midpoint of BC, therefore AD is also the median, angle bisector, and perpendicular bisector of BC. Since AD is the median, BD = CD. Since AD is an altitude, AD is perpendicular to BC. Therefore ADB=ADC=90\angle ADB = \angle ADC = 90^{\circ}. Since AD is the angle bisector, BAD=CAD\angle BAD = \angle CAD. Let BAC=x\angle BAC = x. Then BAD=CAD=x2\angle BAD = \angle CAD = \frac{x}{2}. Consider ABD\triangle ABD. We know ABD=60\angle ABD = 60^{\circ} and ADB=90\angle ADB = 90^{\circ}. Therefore BAD+ABD+ADB=180\angle BAD + \angle ABD + \angle ADB = 180^{\circ} x2+60+90=180\frac{x}{2} + 60^{\circ} + 90^{\circ} = 180^{\circ} x2+150=180\frac{x}{2} + 150^{\circ} = 180^{\circ} x2=30\frac{x}{2} = 30^{\circ} x=60x = 60^{\circ}

\\(\boxed{\angle BAC = 60^{\circ}}\\)

\\(\boxed{\angle BAC = 60^{\circ}}\\)

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