Question

In given figure $P$ and $Q$ are the mid-points of the sides $CA$ and $CB$ respectively of a $△ABC$, right angled at C. Prove that :

(i) $4{AQ}^2=4{AC}^2+{BC}^2$

(ii)$4{BP}^2=4{BC}^2+{AC}^2$

(iii)$4({AQ}^2+{BP}^2)=5{AB}^2$

Answer 1

(i) Since $△AQC$ is a right-angled triangle at C.

By pythagoras theorem

$\therefore$ ${AQ}^2={AC}^2+{QC}^2$

$\Rightarrow$ $4{AQ}^2=4{AC}^2+4{QC}^2$ [Multiplying both sides by 4]

$\Rightarrow$ $4{AQ}^2=4{AC}^2+{\left(2QC\right)}^2$

$\Rightarrow$ $4{AQ}^2=4{AC}^2+{BC}^2$ [$\because BC=2QC$] $\left[Henceproved\right]$

(ii) Since $△BPC$ is a right-angled triangle at C.

By pythagoras theorem

$\therefore$ ${BP}^2={BC}^2+{CP}^2$

$\Rightarrow$ ${4BP}^2=4{BC}^2+{4CP}^2$ [Multiplying both sides by 4]

$\Rightarrow$ ${4BP}^2=4{BC}^2+{\left(2CP\right)}^2$

$\Rightarrow$ ${4BP}^2=4{BC}^2+{AC}^2$ [$\because AC=2CP$] $\left[Henceproved\right]$

(iii) From (i) and (ii), we have

${4AQ}^2={4AC}^2+{BC}^2$ and ${4BP}^2={4BC}^2+{AC}^2$

$\therefore$ ${4AQ}^2+{4BP}^2=({4AC}^2+{BC}^2)+({4BC}^2+{AC}^2)$

$\Rightarrow$ $4({AQ}^2+{BP}^2)=5({AC}^2+{BC}^2)$

$\Rightarrow$ $4({AQ}^2+{BP}^2)=5{AB}^2$ [In $△ABC$, we have ${AB}^2={AC}^2+{BC}^2$] $\left[Henceproved\right]$