(Q16)'O' is any 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.

Given :∆ABC; O' is an interior point of ∆ABC.

OD ⊥ BC, OE ⊥ AC, OF ⊥ AB.

R.T.P :

i) OA² + OB² + OC² - OD² - OE² - OF² = AF² + BD² + CE²

Proof : In OAF, OA² = AF² + OF² [Pythagoras theorem]

⇒ - = AF².....(1)

In ∆OBD,

OB² = BD² + OD²

⇒ - = BD².....(2)

In ∆OCE, OC² = CE² + OE²

- = CE².....(3)

Adding (1), (2) and (3) we get,

² - OF² + ² - OD² + OC² - ² = AF² + ² + CE²

OA² + ² + OC² - ² - OE² - ² = AF² + ² + CE².....(4)

ii) AF² + BD² + CE² = AE² + CD² + BF²

In ∆OAE,

OA² = AE² + OF².....(1)

⇒ ² - ² = AE²

In △OBF, OB² = BF² + OF²

² - ² = BF² .....(2)

In △OCD, OC² = OD² + CD²

² - ² = CD² .....(3)

Adding (1), (2) and (3) we get

OA² - ² + OB² - ² + OC² - ² = AE² + ² + CD²

OA² + ² + OC² - ² - OE² - ² = AE² + ² + BF²

AF² + ² + CE² = ² + CD² + ² [From problem (i)]