This Problem Is Gold

I supplied two methods for solving this problem. The second method uses coordinate geometry.

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Method $1:$

Let $\angle{DAP} = \theta \implies \angle{BAP} = 90 - \theta$ and $\angle{ABD} = 45^{\circ} \implies \angle{BPA} = 45 + \theta \implies \angle{QPD} = 45 + \theta$ and $\angle{CDP} = 45^{\circ} \implies \angle{DQP} = 90 - \theta$

$\therefore \triangle{ABP}$ ~ $\triangle{DPQ}$

Let the height of $\triangle{ABP} = x \implies$ the height of $\triangle{DPQ} = a -x \implies \dfrac{a}{1} = \dfrac{x}{a - x} \implies x = \dfrac{a^2}{a + 1} \implies$

$A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2} \implies a(a^2 - a - 1) = 0$ , since $a > 0 \implies a = \boxed{\dfrac{1 + \sqrt{5}}{2}} = \phi$ .

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Method $2:$

Let $A: (0,0)$ , $B: (0,a)$ , $C: (a,a)$ , $D: (a,0)$ , and $Q: (a,1)$ .

For $BD: y = a - x$

For $AQ: y = \dfrac{x}{a}$

Solving the system above we obtain: $x = \dfrac{a^2}{a + 1} \implies A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2} \implies a(a^2 - a - 1) = 0$ , since $a > 0 \implies a = \boxed{\dfrac{1 + \sqrt{5}}{2}} = \phi$ .