Straight Lines

Let $\vec{AC}=\mathbf{c}$ , $\vec{BD}=\mathbf{d}$ , $\vec{AB}=\vec{DC}=\mathbf{a}$ and $\vec{AD}=\vec{BC}=\mathbf{b}$ . 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 $\lVert \mathbf{d} \rVert = \dfrac{20}{3}$ .

So, we have the diagonals: $\mathbf{c}=\left(\dfrac{8}{\sqrt{5}},-\dfrac{4}{\sqrt{5}}\right)$ and $\mathbf{d}=\left(-\dfrac{20}{3\sqrt{5}},\dfrac{40}{3\sqrt{5}}\right)$ .

We know that, in a parallelogram, $\mathbf{a}+\mathbf{b}=\mathbf{c}$ and $\mathbf{b}-\mathbf{a}=\mathbf{d}$ , so $\mathbf{a}=\dfrac{1}{2}(\mathbf{c}-\mathbf{d})$ and $\mathbf{b}=\dfrac{1}{2}(\mathbf{c}+\mathbf{d})$ .

Substituting the values of the diagonals, we get $\mathbf{a}=\left(\dfrac{22}{3\sqrt{5}},-\dfrac{26}{3\sqrt{5}}\right)$ , and finally $\lVert \mathbf{a} \rVert = \boxed{\dfrac{2\sqrt{58}}{3}}$ .

Relevant wiki: Pythagorean Theorem

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 $\dfrac {BP}{MP} = \tan \theta = \dfrac 34$ , $\implies MP = \dfrac 43 BP = \dfrac 83$ . And by Pythagorean theorem, we have:

$\begin{aligned} AB^2 & = AP^2 + BP^2 \\ & = (AM+MP)^2 + BP*2 \\ & = \left(2+\frac 83 \right)^2 + 2^2 \\ & = \frac {232}9 \\ \implies AB & = \boxed{\dfrac{2\sqrt{58}}3} \end{aligned}$