If the median drawn on the base of a triangle is half its base, the triangle will be:

Let the triangle be $ABC$, where $BC$ is the base and $AD$ is the median drawn to $BC$.
By definition of a median, $D$ is the midpoint of $BC$, so:

$$BD=CD=\frac{1}{2}BC$$

We are given that the median length is equal to half the base:

$$AD=\frac{1}{2}BC$$

Therefore, $AD=BD=CD$.

* In $\Delta ABD$, since $AD=BD$, the opposite angles are equal: $\angle BAD=\angle ABD$.
* In $\Delta ACD$, since $AD=CD$, the opposite angles are equal: $\angle CAD=\angle ACD$.

The full vertex angle $\angle A$ is:

$$\angle A=\angle BAD+\angle CAD=\angle ABD+\angle ACD=\angle B+\angle C$$

We know the sum of angles inside any triangle is $180^{\circ }$:

$$\angle A+\angle B+\angle C=180^{\circ }$$

$$\angle A+\angle A=180^{\circ }⟹2\angle A=180^{\circ }⟹\angle A=90^{\circ }$$

Since one internal angle is $90^{\circ }$, it is a right-angled triangle.