One way to solve this (not necessarily the best way) is to use graphical geometry. In a Cartesian frame of reference let C be (0,0), CB lies along the y-axis, and CD along the x-axis, then let:
A=(a₁,a₂), B=(0,b₂), D=(d₁,0), so C'=(½d₁,0), where b₂>a₂ and d₁>a₁.
CD is y=0 (x-axis), perpendicular bisector of CD is x=d₁/2.
Slope of AB=(a₂-b₂)/a₁, slope of perpendicular is a₁/(b₂-a₂). Because b₂>a₂, the slope of AB is negative, and the slope of the perpendicular is positive.
AB is y=(a₂-b₂)x/a₁+b₂,
perpendicular bisector of AB is y=a₁(x-a₁/2)/(b₂-a₂)+(a₂+b₂)/2.
P is the intersection of the two perpendicular bisectors:
Slope of DP is
s₁=(a₁(d₁/2-a₁/2)/(b₂-a₂)+(a₂+b₂)/2)/(-d₁/2), so s₁<0.
Slope of AD is s₂=a₂/(a₁-d₁), so s₂<0 because d₁>a₁.
Both slopes are negative. If |s₁|>|s₂| then A is to the left of DP, implying that DP does not meet (intersect) AB.
Now, let’s look at the Cosine Rule and apply it to triangle ABD, by joining B and D: BD²=AB²+AD²-2AB.ADcosDAB.
If DAB is obtuse, cosDAB<0, implying BD²>AB²+AD².
BD²=BC²+CD²=b₂²+d₁²; AB²=a₁²+(b₂-a₂)²; AD²=(d₁-a₁)²+a₂².
a₁(d₁-a₁)+a₂(b₂-a₂)>0, which we know to be true from given initial conditions (definition of the vertices A, B, C, D). In other words, when d₁>a₁ and b₂>a₂, angle DAB is always obtuse. For P to lie above AB, that is, above the quadrilateral ABCD (as shown in the given figure), where C is the origin of the frame of reference and BC is along the positive y-axis, while AD is along the x-axis, it follows that b₂>a₂.
The slope of BD is -b₂/d₁ and the slope of AB is -(b₂-a₂)/a₁. If A is to be located above diagonal BD so that ABCD is a convex quadrilateral, as shown in the given figure, then (b₂-a₂)/a₁<b₂/d₁.
If DAB is acute, cosDAB>0, implying BD²<AB²+AD², so a₁(d₁-a₁)+a₂(b₂-a₂)<0, that is, a₂(b₂-a₂)<a₁(a₁-d₁). This time, a₁-d₁ is positive implying A is to the right of D, meaning that DP intersects AB. This can be written (b₂-a₂)/a₁<a₁-d₁.
The implication is that for DAB to be obtuse, A must be located to the left of D (a₁<d₁) so that DP cannot meet AB.