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

If $\angle ABD$ is 60 degrees, what is $\angle BAC$? Given $\angle ABD = 60^{\circ}$ Since B, D, and C are collinear, $\angle ABC = 180^{\circ}$. Thus, $\angle DBC = 180^{\circ} - \angle ABD = 180^{\circ} - 60^{\circ} = 120^{\circ}$ In $\triangle ABC$, 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 $\angle ADB = \angle ADC = 90^{\circ}$. Since AD is the angle bisector, $\angle BAD = \angle CAD$. Let $\angle BAC = x$. Then $\angle BAD = \angle CAD = \frac{x}{2}$. Consider $\triangle ABD$. We know $\angle ABD = 60^{\circ}$ and $\angle ADB = 90^{\circ}$. Therefore $\angle BAD + \angle ABD + \angle ADB = 180^{\circ}$ $\frac{x}{2} + 60^{\circ} + 90^{\circ} = 180^{\circ}$ $\frac{x}{2} + 150^{\circ} = 180^{\circ}$ $\frac{x}{2} = 30^{\circ}$ $x = 60^{\circ}$

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

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