Geometric Inequality Problem: Prove AD < AC in Triangle ABC

To determine if the inequality $AD < AC$ is true, we first consider the configuration given in the problem.

1. The diagram shows triangle $ABC$, with $A$ being the vertex and $B$ and $C$ lying on the line $BC$, which is the base of the triangle. The vertex $D$ is placed such that it seems to be an interior point of triangle $ABC$.

2. Observe that point $D$ lies on the segment connecting the midpoint of the line $BC$ to vertex $A$. The vertex $C$ is at the other end of the line $BC$.

3. Geometrically, since point $D$ lies inside triangle $ABC$ and not on the boundary of triangle $ABC$, segment $AD$ being inside implies it is shorter than segment $AC$, which extends to the triangular boundary on line $BC$.

4. Therefore, the position of $D$ suggests that $AD$ (an interior segment) is indeed shorter than $AC$ (an exterior segment extending from $A$ to $C$). Hence the inequality $AD < AC$ is valid.

Therefore, the given statement $AD < AC$ is True.