Let PQR be a triangle. The points A, B and C are on the sides QR,...

Let the position vector of $P$ is $\overset{\to}{O}$, $Q$ is $\overset{\to}{q}$ and $R$ is $\overset{\to}{r}$.

Now, position vector of $A$ is

$\dfrac{2\overset{\to}{q} + \overset{\to}{r}}{3}$

Position vector of $B$ is $\dfrac{2\overset{\to}{r}}{3}$.

Position vector of $C$ is $\overset{\to}{q}3$.

$\overset{\to}{AB} = \dfrac{\overset{\to}{r} - 2\overset{\to}{q}}{3}$

$\overset{\to}{AC} = \dfrac{- \overset{\to}{r} - \overset{\to}{q}}{3}$

So,

$\overset{\to}{AB} \times \overset{\to}{AC} = \dfrac{1}{3}\left[\left(\overset{\to}{r} - 2\overset{\to}{q}\right) \times \left(- \overset{\to}{r} - \overset{\to}{q}\right)\right]$

$⇒ \overset{\to}{AB} \times \overset{\to}{AC} = \dfrac{1}{3}\left[- \left(\overset{\to}{r} \times \overset{\to}{q}\right) + 2\left(\overset{\to}{q} \times \overset{\to}{r}\right)\right]$

$⇒ \overset{\to}{AB} \times \overset{\to}{AC} = \left(\overset{\to}{q} \times \overset{\to}{r}\right)$

Area of $∆ PQR$ is $= \dfrac{1}{2}\left|\overset{\to}{q} \times \overset{\to}{r}\right|$

Area of $∆ ABC$ is $\dfrac{1}{2}\left|\overset{\to}{AB} \times \overset{\to}{AC}\right|$

Area of $∆ ABC = \dfrac{1}{2}\left|\overset{\to}{q} \times \overset{\to}{r}\right|$

Therefore,

$\dfrac{Area\left(∆ PQR\right)}{Area\left(∆ ABC\right)} = 1$