What's the measure of the sid $BC$ in the quadrilateral below?

For reference:In the ABCD quadrilateral: $AB = CD = 1$ and $AD = 3$. Calculate $BC$, knowing that its diagonals are whole numbers.(Answer:$\frac{8}{3}$)

My progress: It was not mentioned whether the quadrilateral is cyclic or circumscribed.

But: Law of cosines $\triangle ADC$ and $\triangle BAD$ $\implies BD=AC$

$\angle DAC = \angle BDA\\ \angle ABD = \angle ACD$

By triangle inequality:

$3+1<AC<3-1 \therefore AC=3$

Can I say that the quadrilateral is cyclic? By which property? Because If yes I can get the solution

TH.Ptholomeu:

$AB.CD +AD.BC = AC.BD \implies:1+3BC=AC^2\\ 1+3BC=9 \implies \boxed{BC=\frac{8}{3}}$

As you obtained, given the length of $AC$ is whole number, it must be $3$ as $AD - CD \lt AC \lt AD + CD$. Similarly $BD$ must be $3$.

Now coming to your question as to why $ABCD$ is cyclic -

$\triangle ADB \cong \triangle CAD$ and both triangles are isosceles.

So, $\angle ACD = \angle ABD$. As segment $AD$ subtends equal angle on points $B$ and $C$ and both the points are on the same side of $AD$, points $A, B, C $ and $D$ must be cyclic.

So as you showed, applying Ptolemy gets us $ ~ BC = \dfrac 83$.