Question

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

Answer 1

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the ΔARO and applying Pythagoras theorem we get,

AQ² = AR² + OR²

AR² = AQ² - OR²            ...(i)

Similarly, from triangles, BPO, COQ, AOQ, CPO, and BRO we get the following results,

BP² = BO² - OP²          ...(ii)

CQ² = OC² - OQ²        ...(iii)

AQ² = AO² - OQ²        ...(iv)

CP² = OC² - OP²          ...(v)

BR² = OB² - OR²          ...(vi)

Adding (i), (ii) and (iii), we get 

AR² + BP² + CQ² = AQ² - OR² + BO² - OP² + OC² - OQ²    ...(vii)

Adding (iv), (v) and (vi), we get,

AQ² + CP² + BR² = AO² - OR² + BO² - OP² + OC² - OQ²     ...(viii)

From (vii) and (viii), we get,

AR² + BP² + CQ² = AQ² + CP² + BR²

Hence, proved.