In the above figure, O is a point in the interior of a triangle ABC, OD⊥BC, OE⊥AC and OF⊥AB. Show that: (i) OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE² (ii) AF² + BD² + CE² = AE² + CD² + BF²

Construction: Join AO, BO and OC
(i) In △AOE, By Pythagoras Theorem, AO² = AE² + OE² (1)
In △AOF, By Pythagoras Theorem, AO² = AF² + FO² (2)
In △FBO, By Pythagoras Theorem, BO² = BF² + FO² (3)
In △BDO, By Pythagoras Theorem, BO² = BD² + OD² (4)
In △DOC, By Pythagoras Theorem, OC² = OD² + DC² (5)
In △OCE, By Pythagoras Theorem, OC² = OE² + EC² (6)
Add equations (2), (4) and (6) then,
AO² + BO² + OC² = AF² + FO² + BD² + OD² + OE² + EC²
⇒ AO² + BO² + OC² - OD² - OE² - OF² = AF² + BD² + EC²
(ii) Subtract equation (1) from (2), we get,
0 = AF² + FO² - AE² - OE²
⇒ AF² + FO² = AE² + OE². (8)
Subtract equation (3) from (4), we get,
0 = BD² + OD² - BF² - FO²
⇒ BD² + OD² = BF² + FO². (9)
Subtract equation (5) from (6), we get,
0 = OE² + EC² - OD² - DC²
⇒ OE² + EC² = OD² + DC²(10)
Add equations (8), (9) and (10) we get,
AF² + BD² + CE² = AE² + CD² + BF²
Hence, proved.