Q7 of 42 Page 220

‘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, Δ ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB,


Need to prove OA2 + OB2 + OC2-OD2-OE2-OF2 = AF2 + BD2 + CE2


⇒ Join point O to A,B and C


(i) ∠AFO = 90°


AO2 = AF2 + OF2


⇒ AF2 = AO2 - OF2 …….eq(1)


Similarly BD2 = BO2-OD2 …..eq(2)


⇒ CE2 = CO2-OE2 …..eq(3)


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


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


(ii) AF2 + BD2 + CE2 = (AO2-OE2) + ( BO2-OF2) + ( CO2-OD2)


= AE2 + CD2 + BF2


Hence, proved


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