## Triangle Median

#### Problem

The line segments joining the vertices of triangle ABC to the midpoints of the opposite edges are called medians and are concurrent at P.

Prove that P splits each median in the ratio 2:1.

#### Solution

Consider the following diagram.

As C" is the midpoint of AB, AC"=C"B, so the area of triangle APC" = area of triangle C"PB = A_{1}. Similarly area BPA" = area A"PC = A_{2} and area CPB" = area B"PA = A_{3}.

As BB" bisects the area of the triangle, A_{3}+2A_{1} = A_{3}+2A_{2}; hence A_{1}=A_{2}.

Let us consider the way that PB splits triangle ABA": triangles ABP and PBA" have the same altitude, and as area ABP = 2 area PBA", it follows that AP = 2(PA"), proving that P splits the median AA" in the ratio 2:1.

Prove that AA", BB", and CC" are concurrent.

Problem ID: 213 (06 Mar 2005) Difficulty: 3 Star