Put $ \angle D =\alpha \Rightarrow \angle B= \pi - \alpha$, $ AB=BC=x$,$ CD=DE=y$ and $ \angle ACE =\theta$. We have: $ [ABCDE ]= [ABC]+[ACE]+[CDE]$. $ [ABC]=\frac{1}{2}\cdot x^2\cdot\sin(\pi -\alpha)$ and $ [CDE]=\frac{1}{2}\cdot y^2 \cdot \sin\alpha$. $ AC= 2x \cos\frac{\alpha}{2}$ and $ CE=2y\sin\frac{\alpha}{2}$. Therefore $ [ACE]=\frac{1}{2}\sin\theta\cdot(2x\cos\frac{\alpha}{2}) \cdot (2y\sin\frac{\alpha}{2}) = xy\sin \theta \cdot\sin\alpha$. Hence $ [ABCDE]=\frac{\sin\alpha}{2}\cdot(x^2+y^2+2xy\sin\theta)$. But $ \angle BCD = \frac{\pi}{2}+\theta$, so : $ [ABCDE]= \frac{\sin\alpha}{2}\cdot(x^2+y^2-2xy\cos\angle BCD)=\frac{BD^2}{2}\cdot \sin \alpha$. So we have the equation: $ BD^2=\frac{2\sqrt{2}}{\sin \alpha}$ a) if $ \angle B = \frac{3\pi}{4} \Rightarrow \alpha=\frac{\pi}{4}\Rightarrow BD=2$. b) $ BD$ is minimal if $ \frac{2\sqrt{2}}{\sin \alpha}$ is minimal, i.e. if $ \alpha=\frac{\pi}{2}$. In this case $ BD=\sqrt{2\sqrt{2}}$.