$AB=AE=s$ and $BC=CD=t$. $AC^{2}=s^{2}+t^{2}+2st \cos \alpha=1$. Area $\triangle ABE=\frac{1}{2}s^{2}$, area $\triangle BCD=\frac{1}{2}t^{2}$. $\triangle BED\ :\ BE=\sqrt{2}s,BE=\sqrt{2}t,DF=\sqrt{2}t \cos \alpha$, area $\triangle BED\ =\frac{1}{2} \cdot \sqrt{2}s \cdot \sqrt{2}t \cos \alpha=\frac{1}{2}(2st\cos \alpha)$. Area $ABCDE =\frac{1}{2}s^{2}+\frac{1}{2}t^{2}+\frac{1}{2}(2st\cos \alpha)=\frac{1}{2}$.

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Here is a nice Synthetic Solution. Let $AK$ be a line $\perp$ to $AC$ and $AK=AC$. Now, we know $AB=AE$ and as $\angle CAK=\angle BAE\implies\angle KAE=\angle CAB$ and $AC=AK$. So, $\triangle CAB\cong\triangle KAE$. So, $[CAB]=[KAE]$. Now, $EK=BC=CD$. Now let $\angle AKE=\theta\implies\angle EKC=45^\circ-\theta$. So, as $\angle BCA=\angle AKE=\theta\implies\angle KCD=45^\circ-\theta$. So, $EK\|CD\implies\angle KED=\angle CDE$. If $KC\cap ED=P$ then $\triangle CPD\cong\triangle KPE\implies [CPD]=[KPE]$. So, $[ABCDE]=[ABC]+[AEPC]+[CPD]=[AEK]+[AEPC]+[KEP]=[AKC]=\frac{1}{2}$.

One of the official solutions: Since $\angle B + \angle D + \angle E = 360^\circ$, we can properly divide $\triangle ABC$ in $\angle B$ in two pieces and rotate this parts to form right-angled trapezoid like on diagram. The height and sum of the bases of this trapezoid are equal to $AC$.

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