Bang the box

There is a property that the total force on an area inside any fluid is equal to the $\text{pressure applied by that fluid at the center of the area}$ $\times$ $\text{total area}$

now let $area of \Delta ABC = area \Delta ADC = \Delta$

and when we calculate the depth of the center of both the triangle we will sea that the depth of the center of $\Delta ADC$ is twice the depth of the center of $\Delta ABC$

so $\dfrac{F_1}{F_2}$ $=$ $\dfrac{\delta\times g\times h \times \Delta}{\delta\times g\times 2h\times \Delta}$ $=$ $0.5$

where $\delta = density~ given~ liquid$

$h = depth ~of~ center~ of \Delta ABC$

$g = acceleration ~due ~to ~gravity$

Being triangular sections center of pressure of the upper wedge is located at 1/3 from the surface meanwhile the center of the lower wedge is at 2/3, so the ratio is 0.5