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 &#x2206 OAR, we can write:
AR2 = OA2 – OR2 —— Eqn. (I)
For &#x2206 OBR, we can write:
BP2 = OB2 – OR2 —— Eqn. (II)
For &#x2206 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

READ :  AB = BC and AD is perpendicular to CD. Prove that AC2 = 2.BC. DC