It is enough to measure $BD$ because $(ABCD) = \frac{1}{2} BD^2$.
Indeed, extend $AD$ to meet $BC$ at $P$. Since $\angle ABP = \angle BAP = 45^\circ$, we have $AP = BP$ and $\angle APB = 90^\circ$. Hence triangle $ABP$ is right and isosceles, so that $(ABP) = \frac{1}{2} BP^2$.

Next, triangle $CDP$ has $\angle PCD = 45^\circ$, $\angle CPD = 90^\circ$. Therefore it is right and isosceles too, with
$$(CDP) = \frac{1}{2} DP^2. \text{ It follows that}$$
$$(ABCD) = (ABP) + (CDP) = \frac{1}{2}(BP^2 + DP^2) = \frac{1}{2} BD^2.$$
The last equality follows from Pythagoras' theorem in triangle $BDP$.