Problem Analysis: We are given a cyclic quadrilateral ABCD where the side AB is the diameter of the circumscribing circle. We know that $\angle ADC = 129^\circ$ and need to find the measure of $\angle BAC$. We will use properties of cyclic quadrilaterals and angles in a semicircle.

Cyclic Quadrilateral Angle Properties

In a cyclic quadrilateral, opposite angles are supplementary (add up to $180^\circ$). Therefore, for quadrilateral ABCD:

• $\angle ADC + \angle ABC = 180^\circ$

Substitute the given value:

• $129^\circ + \angle ABC = 180^\circ$
• $\angle ABC = 180^\circ - 129^\circ$
• $\angle ABC = 51^\circ$

Angle in a Semicircle

Since AB is the diameter of the circle, the angle subtended by the diameter at any point on the circumference is a right angle ($90^\circ$). Therefore, for point C on the circumference:

• $\angle ACB = 90^\circ$

Finding Angle BAC in Triangle ABC

Now consider the triangle $\triangle ABC$. The sum of the angles in any triangle is $180^\circ$. We know $\angle ABC$ and $\angle ACB$. We can find $\angle BAC$:

• $\angle BAC + \angle ABC + \angle ACB = 180^\circ$
• $\angle BAC + 51^\circ + 90^\circ = 180^\circ$
• $\angle BAC + 141^\circ = 180^\circ$
• $\angle BAC = 180^\circ - 141^\circ$
• $\angle BAC = 39^\circ$

Conclusion

The measure of $\angle BAC$ is $39^\circ$. This corresponds to Option A.