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Given : A triangle ABC, in which OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.
To Prove:
(i) OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE²
(ii) AF² + BD² + CE² = AE² + CD² + BF²
Const: Join O-B, O-C, and O-A.
Proof: In right triangle OFA, we have
OA² = AF² + OF²    ...(i)
[Using Pythagoras theorem]
In right triangle OBD, we have
OB² = OD² + BD²    ...(ii)
[Using Pythagoras theorem]
In right triangle OEC, we have
OC² = OE² + CE²    ...(iii)
[Using Pythagoras theorem]
Adding (i), (ii) and (iii), we get
OA² + OB² + OC² = AF² + OF² + OD² + BD² + OE² + CE²
⇒ OA² + OB² + OC² = AF² + BD² + CE² + OD² + OF² + OE²
⇒ AF² + BD² + CE² = OA² + OB² + OC² - OD² - OF² - OE² Proved.
(ii) We can re-write the above proved result as follows:
AF² + BD² + CE² = (OA² - OE²) + (OB² - OF²) + (OC² - OD²)
= AE² + CD² + BF²
Hence, AF² + BD² + CE² = AE² + CD² + BF².