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 = AE² + CD² + BF²

Answer 1

In right triangles ODB and ODC, we have
OB² = OD² + BD²
OC² = OD² + CD²
∴ OB² - OC² = (OD² + BD²) - (OD² + CD²)
⇒ OB² - OC² = BD² - CD²  ....(i)
Similarly, we have
OC² - OA² = CE² - AE²      ....(ii)
OA² - OB² = AF² - BF²      ....(iii)
Adding (i), (ii) and (iii), we get
(OB² - OC²) + (OC² - OA²) + (OA² - OB²) = (BD² - CD²) + (CE² - AE²) + (AF² - BF²)
⇒ (BD² + CE² + AF²) - (AE² + CD² + BF²) = 0
⇒ AF² + BD² + CE² = AE² + CD² + BF².