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Answer: $\frac{1}{8}(1+\sqrt{2})$. Reflect the quadrilateral in line $AB$, and reflect the resulting pentagon again in line $AD$; see Figure 1. This produces an octagon of fixed perimeter $$4(BC+CD)=4$$ and area four times that of $ABCD$. The maximal area is obtained for a regular octagon of edge length $\frac{1}{2}$. Since the area of a regular octagon of edge $a$ is known (or easily verified) to be $2(1+\sqrt{2})a^2$, the maximal area of the original quadrilateral is $\frac{1}{4}\cdot 2(1+\sqrt{2}){\left(\frac{1}{2}\right)}^2=\frac{1}{8}(1+\sqrt{2})$.



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Alternative solution.

Let us fix the segment $BD$, and consider the points $A$ and $C$ as variables. The locus of points $A$ fulfilling $\angle A=90^{\circ }$ is a (semi-)circle with diameter $BD$. In order to maximise the area of triangle $BAD$, the length of the altitude from $A$ must be maximised, which occurs when $A$ lies on the perpendicular bisector of $BD$, so that $AB=AD$. The locus of points $C$ fulfilling $BC+CD=1$ is (an arc of) an ellipse with foci $B$ and $D$. Again, in order to maximise the area of triangle $BCD$, the length of the altitude from $C$ must be maximised, which again occurs for $C$ on the perpendicular bisector of $BD$, so that $BC=CD=\frac{1}{2}$. The maximal quadrilateral satisfying the conditions will thus be mirror-symmetric with $$AB=AD\text{and}BC=CD=\frac{1}{2},$$



as in Figure 2. Put $\alpha =\angle BCA=\angle DCA$. Using some trigonometry, we find $BD=\sin \alpha$ and $AB=\frac{1}{\sqrt{2}}\sin \alpha$, so that the area of $ABCD$ can be expressed as $$\begin{array}{rl}|ABCD|& =|BCD|+|BAD|\\ & =\frac{1}{8}\sin 2\alpha +\frac{1}{4}{\sin }^{2}2\alpha \\ & =\frac{1}{8}(\sin 2\alpha +1-\cos 2\alpha )\\ & =\frac{1}{8}\left(1+\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin 2\alpha -\frac{1}{\sqrt{2}}\cos 2\alpha \right)\right)\\ & =\frac{1}{8}\left(1+\sqrt{2}\sin (2\alpha -45^{\circ })\right).\end{array}$$ Since $0^{\circ }\le \alpha \le 90^{\circ }$, this function obtains its unique maximum $\frac{1}{8}(1+\sqrt{2})$ for $\alpha =62.5^{\circ }$.