Exercise 51

In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=5$, $PD=12$, and $DE=13$. What is the area of $AEDC$?

Solution

Since $5^2+12^2=13^2$, $PD\perp PE$.
The centroid divides each median $2:1$, so $CE=3\cdot PE=15$ and $AD=3\cdot PD=36$.
Since $AD\perp CE$, $[AEDC]=\frac12\cdot AD\cdot CE=\frac12(36)(15)=\boxed{270}$ by theorem 16.