Let $a, b, c, d$ be the lengths of the tangent line segments at vertices $A, B, C, D$, respectively, and $\alpha = \angle APB$ (Fig. 25). We have $a + b > b + c$ and $a + b > a + d$ as $AB$ is the longest side, implying $a > c$ and $b > d$. By the law of cosines:
$$(a+b)^2 = PA^2 + PB^2 - 2 \cdot PA \cdot PB \cdot \cos \alpha,$$
$$(b+c)^2 = PB^2 + PC^2 + 2 \cdot PB \cdot PC \cdot \cos \alpha,$$
$$(c+d)^2 = PC^2 + PD^2 - 2 \cdot PC \cdot PD \cdot \cos \alpha,$$
$$(d+a)^2 = PD^2 + PA^2 + 2 \cdot PD \cdot PA \cdot \cos \alpha.$$

Fig. 25

$2(ab + cd - bc - da) = -2(PA \cdot PB + PB \cdot PC + PC \cdot PD + PD \cdot PA) \cos \alpha$. The l.h.s. can be expressed as $2(a - c)(b - d)$ which is positive since $a > c$ and $b > d$. The parenthesized expression in the r.h.s. is also positive. Hence $\cos \alpha$ must be negative. This means that the angle $APB$ is obtuse.