A median splits a triangle into two equal-area parts

Text

Let $PQR$ be any triangle and let $M$ be the midpoint of side $QR$. Prove that the median $PM$ splits the triangle into two equal-area triangles, that is, $PQM$ and $PMR$ have the same area.

Solution

Since $M$ is the midpoint of $QR$, the two segments $QM$ and $MR$ have the same length: $\overline{QM}=\overline{MR}=\tfrac12\,\overline{QR}.$ Take $QM$ as the base of triangle $PQM$ and $MR$ as the base of triangle $PMR$. Because $Q$, $M$, $R$ are collinear, the two triangles have the same height: the distance $h$ from vertex $P$ to the line $QR$. Their areas are then $\mathcal{A}(PQM)=\frac{\overline{QM}\cdot h}{2},\qquad \mathcal{A}(PMR)=\frac{\overline{MR}\cdot h}{2}.$ Since $\overline{QM}=\overline{MR}$, the two areas coincide: $\mathcal{A}(PQM)=\mathcal{A}(PMR)=\tfrac12\,\mathcal{A}(PQR).$ The two triangles are therefore equivalent. $\;\blacksquare$