It's A Square Problem!

Let the length of each side of the square be $2a$ and the angle $\angle {FBC}$ be $α$ . Let the center of the semicircle be $O$ . Then $\angle {DOF}=45\degree-\dfrac{α}{2}$ . $|\overline {BE}|=|\overline {AB}|=2a,|\overline {EF}|=|\overline {FD}|$ . Now $\tan (45\degree-\dfrac{α}{2})=\dfrac{|\overline {FD}|}{|\overline {OD}|}\implies \tan (\dfrac{α}{2})=\dfrac{|\overline {AB}|-2|\overline {FD}|}{|\overline {AB}|+2|\overline {FD}|}$ , and $\tan α=\dfrac{|\overline {CF}|}{|\overline {AB}|}=1-\dfrac{|\overline {FD}|}{|\overline {AB}|}$ . From these we get $3\tan^3 (\dfrac{α}{2})+5\tan^2 (\dfrac{α}{2})+\tan (\dfrac{α}{2})-1=0\implies \tan (\dfrac{α}{2})=\dfrac{1}{3}\implies \tan (45\degree-\dfrac{α}{2})=\dfrac{1}{2}\implies |\overline {EF}|=|\overline {FD}|=\dfrac{a}{2}\implies |\overline {BF}|=|\overline {BE}|+|\overline {EF}|=2a+\dfrac{a}{2}=\dfrac{5a}{2}$ . Hence $\dfrac{|\overline {BF}|}{|\overline {AB}|}=\dfrac{5a}{2\times 2a}=\dfrac{5}{4}=\boxed {1.25}$

Let $a$ be the length of a side of the square $ABCD$ .

$(x - \dfrac{a}{2})^2 + y^2 = \dfrac{a^2}{2} \implies \dfrac{(2x - a)^2}{4} + y^2 = \dfrac{a^2}{4} \implies$

$x^2 - ax + y^2 = 0 \implies y = \sqrt{x(a - x)}$

$m_{BE} = \dfrac{\sqrt{x(a - x)} - a}{x} \implies m_{\perp} = m_{PE} =$ $\dfrac{2\sqrt{x(a - x)}}{2x - a} =$

$-\dfrac{x}{\sqrt{x(a - x)} - a} \implies$ $2ax - 2x^2 - 2a\sqrt{x(a - x)} = ax - 2x^2 \implies$

$x = 2\sqrt{x(a - x)} \implies x^2 = 4ax - 4x^2 \implies x(4a - 5x) = 0$ and $x \neq 0$

$\implies x = \dfrac{4a}{5} \implies y = \dfrac{2a}{5}$

Using $E:(\dfrac{4a}{5}, \dfrac{2a}{5})$ and $B:(0,a) \implies m_{BE} = -\dfrac{3}{4} \implies$ $y = -\dfrac{3}{4}x + a$

Using line $x = a \implies y = \dfrac{a}{4}$ and using $F:(a,\dfrac{a}{4})$ and $B:(0,a)$ $\implies$

$\overline{BF} = \sqrt{\dfrac{25a^2}{16}} = \dfrac{5}{4}a \implies$ $\dfrac{\overline{BF}}{AB} = \dfrac{5}{4} = \boxed{1.25}$