If $P$ And $Q$ are the points on side $CA$ and $CB$ respectively of  $\vartriangle ABC$, right angled at C, prove that  $( AQ^{2}+BP^{2})=( AB^{2}+ PQ^{2})$.

Given: $P$ and $Q$ are the points on side $CA$ and $CB$ respectively of  $\vartriangle ABC$, right angled at $C$.

To do: To prove that $( AQ^{2}+BP^{2})=(AB^{2}+ PQ^{2})$ 

Solution:
Using the Pythagoras theorem in $\vartriangle ABC$,

$\vartriangle ACQ$, $\vartriangle BPC$, $\vartriangle PCQ$, we get   

$AB^{2}=AC^{2}+BC^{2}$             ....................$( i)$

$AQ^{2}=AC^{2}+CQ^{2} $          ....................$( ii)$

$BP^{2}=PC^{2}+BC^{2}  $           ................... $( iii)$

$PQ^{2}=PC^{2}+CO^{2} $           ....................$( iv)$

Adding the equations $( ii)$ and $( iii)$ we get

$AQ^{2}+BP^{2}=AC^{2}+CQ^{2}+PC^{2}+BC^{2}$   

$=( AC^{2}+BC^{2})+( CQ^{2}+PC^{2})$

$=AB^{2}+PQ^{2}$

As $L.H.S=AQ^{2}+BP^{2}$

$=AB^{2}+PQ^{2}=R.H.S$

Hence proved.
 

Updated on: 10-Oct-2022

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