In triangle ABC, if line segment BD is both the altitude and the median, we can conclude that triangle ABC is an isosceles triangle with AB equal to AC.
To understand this, let’s first break down what it means for BD to be an altitude and a median. An altitude is a line segment drawn from a vertex to the opposite side, making a right angle with that side. In this case, BD is perpendicular to side AC. A median, on the other hand, is a line segment drawn from a vertex to the midpoint of the opposite side. Hence, D is the midpoint of side AC.
Since BD is both perpendicular to AC (as the altitude) and divides AC into two equal segments (as the median), triangle ABC exhibits symmetrical properties. This symmetry indicates that the two sides AB and AC must be equal in length. Thus, we can infer that triangle ABC is isosceles, with the equal sides being AB and AC, and base AC.
Additionally, the fact that BD is both the altitude and median also indicates that angle A must be bisected by segment BD. This akin to the properties of an isosceles triangle, where the angle bisector (BD) also serves as the median and altitude when connecting the vertex to the base.
In conclusion, when BD acts as both the altitude and median in triangle ABC, it affirms that triangle ABC is indeed isosceles with the sides AB and AC being congruent.