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 = AQ² + CP² + BR².

Join OA, OB and OC.

By formula,

By pythagoras theorem,

(Hypotenuse)² = (Perpendicular)² + (Base)²

In right-angled triangle AOR,

⇒ AO² = AR² + OR²

⇒ AR² = AO² - OR² ……….(1)

In right-angled triangle BOP,

⇒ BO² = BP² + OP²

⇒ BP² = BO² - OP² ……….(2)

In right-angled triangle COQ,

⇒ CO² = CQ² + OQ²

⇒ CQ² = CO² - OQ² ……….(3)

In right-angled triangle AOQ,

⇒ AO² = AQ² + OQ²

⇒ AQ² = AO² - OQ² ……….(4)

In right-angled triangle COP,

⇒ CO² = CP² + OP²

⇒ CP² = CO² - OP² ……….(5)

In right-angled triangle BOR,

⇒ BO² = BR² + OR²

⇒ BR² = BO² - OR² ……….(6)

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

⇒ AR² + BP² + CQ² = AO² - OR² + BO² - OP² + CO² - OQ² ……….(7)

Adding equations (4), (5) and (6), we get :

⇒ AQ² + CP² + BR² = AO² - OQ² + CO² - OP² + BO² - OR²

⇒ AQ² + CP² + BR² = AO² - OR² + BO² - OP² + CO² - OQ² ………..(8)

From equations (7) and (8), we get :

⇒ AR² + BP² + CQ² = AQ² + CP² + BR².

Hence, proved that AR² + BP² + CQ² = AQ² + CP² + BR².