In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that AR^{2} +BP^{2} +CQ^{2} = AQ ^{2} + CP ^{2}+BR^{2}

# solution :

Construction : joinA-0, BO and CO . For ∆ OAR, we can write:

AR^{2} = OA^{2} – OR^{2} —— Eqn. (I)

For ∆ OBR, we can write:

BP^{2} = OB^{2} – OR^{2} —— Eqn. (II)

For ∆ OCQ, we can write:

CQ^{2} = OC^{2} – OQ^{2} —— Eqn. (III)

Adding Eqn. (I), (II) & (III) , We get:

AR^{2} +BP^{2} +CQ^{2} = OA^{2}+OB^{2}+OC^{2}-OR^{2}-OP^{2}-OQ^{2}

= ( OA^{2} – OQ^{2} ) + ( OB^{2} -OR^{2} ) + ( OC^{2} -OP^{2} )

= AQ ^{2} + CP ^{2}+BR^{2} __Proved__