Isosceles Triangle Median: Proving BAD = DAC Angle Equality

To solve this problem, we need to examine the properties of an isosceles triangle where the median is drawn from the vertex to the base.

Given $\triangle ABC$ is isosceles with $AB = AC$, and $AD$ is the median to the base $BC$, it follows certain properties specific to isosceles triangles:

• In an isosceles triangle, the median from the vertex angle is also the altitude and the angle bisector for the angle at the vertex.
• Since $AD$ is the median, $BD = DC$.
• This particular property of isosceles triangles means that $AD$ not only divides the base $BC$ equally but also divides the vertex angle $\angle BAC$ equally, thus forming the angle bisector.

Therefore, the angle $\angle BAD$ is equal to $\angle DAC$ because $AD$ serves as the angle bisector of $\angle BAC$ in the isosceles triangle $\triangle ABC$.

Thus, it is true that $\angle BAD = \angle DAC$ in this triangle.

Yes.