If
\(\dfrac{p}{q}=\dfrac{q}{r}\) ,
then prove the given problem.
\(\begin{array}{|rcll|} \hline \dfrac{p}{q} &=& \dfrac{q}{r} \\ pr &=& q^2 \\ \mathbf{2pr} &=& \mathbf{2q^2} \qquad (1) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline && (p+q+r)(p-q+r) \\ &=& \Big((p+r)+q\Big) \Big((p+r)-q\Big) \\ &=& (p+r)^2-q^2 \\ &=& p^2+2pr+r^2-q^2 \quad &| \quad \mathbf{2pr=2q^2} \\ &=& p^2+2q^2+r^2-q^2 \\ \mathbf{(p+q+r)(p-q+r)} &=& \mathbf{p^2+q^2+r^2}\ \checkmark \\ \hline \end{array}\)