Weird Figure

Using $AB=CD$ and $AD=AX+DX$ , we note that:

$[ABX]+[DCX]=\dfrac{1}{2}\cdot AB \cdot AX+ \dfrac{1}{2}\cdot CD \cdot DX= \dfrac{AB}{2}(AX+DX)=\dfrac{AB\cdot AD}{2} = \dfrac{[ABCD]}{2}$

$\Rightarrow [ABCD]= 2([ABX]+[DCX]) \\ \Rightarrow [ABX]+[DCX]+[BXC] = 2([ABX]+[DCX]) \\ \Rightarrow [BXC] = [ABX]+[DCX] \\ \Rightarrow [BA'X]+\text{red region} = [ABX]+[DCX]$

Since $A'$ is reflection of $A$ in $BX$ we easily have $\Delta BAX \cong \Delta BA'X \Rightarrow [BAX]=[BA'X]$ and using this fact from the above equation we have:

$[BAX]+\text{red region} = [ABX]+[DCX] \\ \Rightarrow \text{red region} =[DCX] = \text{blue region}$

Since they are equal, their ratio is simply $\boxed{1}$ .

• $[ABC]$ denotes area of triangle $ABC$ .