A Special Square

Let the side length of square $ABCD$ be $a$ . We note $\triangle ABP$ and $\triangle DPQ$ are similar. Then $\dfrac {A_{\triangle DPQ}}{A_{\triangle ABP}} = \dfrac {1^2}{a^2} \implies \dfrac {A_{\triangle DPQ}}{\frac a2} = \dfrac {1^2}{a^2} \implies A_{DPQ} = \dfrac 1{2a}$ . And $A_{\triangle ADP} = A_{\triangle ADQ} - A_{\triangle DPQ} = \dfrac a2 - \dfrac 1{2a}$ . Now we have:

$\begin{aligned} A_{\triangle ABD} & A_{\triangle ABP} + A_{\triangle ADP} \\ \frac {a^2}2 & = \frac a2 + \frac a2 - \frac 1{2a} & \small \blue{\text{Multiply both sides by }2a} \\ a^3 & = 2a^2 - 1 \\ a^3 - 2a^2 + 1 & = 0 \\ (a-1)(a^2 - a - 1) & = 0 & \small \blue{\text{Since }a > 1} \\ \implies a & = \frac {1+\sqrt 5}2 \approx \boxed{1.618} & \small \blue{\varphi \text{, the golden ratio.}} \end{aligned}$

I provided two solutions:

Solution using coordinate geometry:

For $BD: y = a - x$ and for $AQ: y = \dfrac{1}{a}x \implies x = \dfrac{a^2}{a + 1} \implies$

$A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$ $\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies$

$\boxed{a = \dfrac{1 + \sqrt{5}}{2} \approx 1.61803}$ .

Solution using similar triangles:

Since vertical angles are congruent and $AB \parallel CD \implies$ alternate interior angles are congruent $\implies \triangle{ABP} \sim \triangle{DPQ} \implies$

$\dfrac{a}{1} = \dfrac{x}{a - x} \implies x = \dfrac{a^2}{a + 1} \implies A_{\triangle{ABP}} = \dfrac{a^3}{2(a + 1)} = \dfrac{a}{2}$

$\implies a(a^2 - a - 1) = 0 \:\ a > 0 \implies \boxed{a = \dfrac{1 + \sqrt{5}}{2} \approx 1.61803}$ .