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OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB of ∆ ABC |Prove that AR2 +BP2 +CQ2 = AQ 2 + CP 2+BR2

In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that AR2 +BP2 +CQ2 = AQ 2 + CP 2+BR2

solution :

Construction : joinA-0, BO and CO . For ∆ OAR, we can write:
AR2 = OA2 - OR2 ------ Eqn. (I)
For ∆ OBR, we can write:
BP2 = OB2 - OR2 ------ Eqn. (II)
For ∆ OCQ, we can write:
CQ2 = OC2 - OQ2 ------ Eqn. (III)
Adding Eqn. (I), (II) & (III) , We get:
AR2 +BP2 +CQ2 = OA2+OB2+OC2-OR2-OP2-OQ2
        = ( OA2 - OQ2 ) + ( OB2 -OR2 ) + ( OC2 -OP2 )
        = AQ 2 + CP 2+BR2    Proved

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