= Solution
{c}
With $t$ as in <1/a/i>, the <foot of a perpendicular> lies strictly between the endpoints precisely when $0<t<1$. Since the <denominator> is positive, this gives
$$
\boxed{0<(\mathbf c-\mathbf a)\cdot(\mathbf b-\mathbf a)<\|\mathbf b-\mathbf a\|^2.}
$$
Equivalently, both <dot products>
$$
(\mathbf c-\mathbf a)\cdot(\mathbf b-\mathbf a)>0,\qquad
(\mathbf c-\mathbf b)\cdot(\mathbf a-\mathbf b)>0
$$
are positive: the <angles> of the <triangle> at $A$ and $B$ are both less than $\pi/2$. If “between” includes the endpoints, replace both strict <inequalities> by non-strict ones; equality gives $P=A$ or $P=B$.
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