In the given figure, $O P, O Q$ and $O R$ are drawn perpendiculars to the sides $B C, C A$ and $A B$ respectively of triangle $A B C$ . Prove that: ${A R}^{2} + {B P}^{2} + {C Q}^{2} = {A Q}^{2} + {C P}^{2} + {B R}^{2}$

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Solution

Consider the diagram shown below.

Consider $△ A O R$ . Using pythagoras theorem, we have
$A R^{2} = O A^{2} - O R^{2}$ ...... (1)

Consider $△ B O P$ . Using pythagoras theorem, we have
$B P^{2} = O B^{2} - O P^{2}$ ...... (2)

Consider $△ C O Q$ . Using pythagoras theorem, we have
$C Q^{2} = O C^{2} - O Q^{2}$ ...... (3)

Add equations (1), (2) and (3).
$A R^{2} + B P^{2} + C Q^{2} = O A^{2} - O R^{2} + O B^{2} - O P^{2} + O C^{2} - O Q^{2}$
$= (O A^{2} - O Q^{2}) + (O C^{2} - O P^{2}) + (O B^{2} - O R^{2})$
$= A Q^{2} + C P^{2} + B R^{2}$
Therefore,
$L H S = R H S$
Hence, the given expression is proved.

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