Question

From a point O in the interior of aΔABC, perpendicular OD, OE and OF are drawn to the sides BC, CA and AB respectively. Prove that: AF² + BD² + CE² = OA² + OB² + OC² - OD² - OE² - OF²

Answer 1

In right triangle OFA, ODB and OEC, we have
OA² = AF² + OF²
OB² = BD² + OD²
OC² = CE² + OE²
Adding all these results, we get
OA² + OB² + OC² = AF² + BD² + CE² + OF² + OD² + OE²
⇒ AF² + BD² + CE² = OA² + OB² + OC² - OD² - OE² - OF².