Straight Lines

Let the diagonals $AC$ and $BD$ intersect at $M$ and $\angle BMC = \theta$ . We note that

$\begin{cases} AC: x+2y - 3 = 0 & \implies y = - \dfrac 12 x + \dfrac 32 & ...(1) \\ BD: 2x+y - 3 = 0 & \implies y = - 2 x + 3 & ...(2) \end{cases}$

Therefore the gradients of diagonals $AC$ and $BD$ are $\tan^{-1} - \frac 12$ and $\tan^{-1} - 2$ respectively and that

$\begin{aligned} \theta & = \tan^{-1} - \frac 12 - \tan^{-1} - 2 \\ \tan \theta & = \frac {\left(- \frac 12\right)-(-2)}{1+\left(- \frac 12\right)(-2)} = \frac 34 \end{aligned}$

Now, let the perpendicular from point $B$ to $AC$ extension be $BP$ . Given that $AC=4$ and $[ABCD]=8$ , $BP = 2$ . Since $\dfrac {BP}{BM} = \sin \theta = \dfrac 35$ , $\implies BM = \dfrac 53 BP = \dfrac {10}3$ and $BD = 2 BM = \boxed{\dfrac {20}3}$ .

Let $\vec{AC}=\mathbf{c}$ and $\vec{BD}=\mathbf{d}$ . One parallel vector to $\vec{AC}$ is $(2,-1)$ and to $\vec{BD}$ is $(-1,2)$ . So, $\hat{\mathbf{c}}=\left(\dfrac{2}{\sqrt{5}},-\dfrac{1}{\sqrt{5}}\right)$ and $\hat{\mathbf{d}}=\left(-\dfrac{1}{\sqrt{5}},\dfrac{2}{\sqrt{5}}\right)$ .

We know that $\dfrac{1}{2}\lVert \mathbf{c} \times \mathbf{d} \rVert = 8$ , so $\lVert \mathbf{c} \rVert \lVert \mathbf{d} \rVert \lVert \hat{\mathbf{c}} \times \hat{\mathbf{d}} \rVert = 16$ . But $\lVert \mathbf{c} \rVert = 4$ and $\lVert \hat{\mathbf{c}} \times \hat{\mathbf{d}} \rVert = \dfrac{3}{5}$ , so $\lVert \mathbf{d} \rVert = \boxed{\dfrac{20}{3}}$ .