Given that the perpendicular bisector of DC and AB meets at P, and DAB is an obtuse angle. How to prove that DP will never meet AB?

reopened

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:

(d₁/2,a₁(d₁/2-a₁/2)/(b₂-a₂)+(a₂+b₂)/2).

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.

Therefore, b₂²+d₁²>a₁²+(b₂-a₂)²+(d₁-a₁)²+a₂².

Expanding:

b₂²+d₁²>a₁²+b₂²-2a₂b₂+a₂²+d₁²-2a₁d₁+a₁²+a₂²,

0>2a₁²-2a₂b₂-2a₁d₁+2a₂²,

0>a₁²-a₂b₂-a₁d₁+a₂²,

0>-a₁(d₁-a₁)-a₂(b₂-a₂),

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.

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