A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. In triangle ABC, we have three medians: AD connects vertex A to the midpoint D of side BC, BE connects vertex B to the midpoint E of side AC, and CF connects vertex C to the midpoint F of side AB. We will prove that all three medians intersect at a single point called the centroid.
Let's begin the proof. First, consider two medians AD and BE. Since they are line segments within the triangle starting from different vertices, they must intersect at some point. Let's call this intersection point G. Next, we connect the midpoints D and E. By the Midsegment Theorem, the segment DE is parallel to side AB and has length equal to half of AB.
Now we can establish similarity between triangles. Since AB is parallel to DE, triangles ABG and EDG are similar by angle-angle-angle similarity. The corresponding angles are equal due to parallel lines and transversals. Since DE equals half of AB, the ratio AB to DE equals 2. From the similarity, we get that AG to GD equals BG to GE equals 2. This means point G divides each median in a 2 to 1 ratio from the vertex.
Now let's consider the third median. We examine medians AD and CF and connect midpoints D and F. By the Midsegment Theorem, DF is parallel to AC and DF equals half of AC. This creates similar triangles ACG and FDG. Using the same reasoning as before, their intersection point also divides the medians in a 2 to 1 ratio from the vertex. Since there can only be one point on median AD that divides it in a 2 to 1 ratio, this must be the same point G.
We have successfully proven that all three medians of a triangle intersect at a single point. This special point is called the centroid of the triangle. The centroid has the remarkable property that it divides each median in a 2 to 1 ratio, with the longer segment being from the vertex to the centroid. This fundamental theorem holds for any triangle, making the centroid one of the most important points in triangle geometry.