As the picture shows, there is a circle on the plane of radius $R$ centered at $A$. I found that if a line segment $PQ$ of length $d\ (0<d\leq R)$ intersecting with $AB$ satisfies that the endpoint $P$ is outside the circle, then $|QB|<d$.
Is there any simple and clear proof for it?
$P$ being outside the circle means that $|AP| > R$. Let's call $S$ the point of intersection of $AB$ and $PQ$. Consider the triangle $ASP$: by the triangle inequality, we get
$$|AP| \leq |AS| + |SP|$$
Analogously, for the triangle $QSB$ we get
$$|QB| \leq |QS| + |SB|$$
Add this two inequalities to get
$$|AP| + |QB| \leq |AS| + |SP| + |QS| + |SB|$$
Now, $AS$ and $SB$ together form the segment $AB$, thus $|AS| + |SB| = |AB| = R$, and similarly $|QS| + |SP| = |QP| = d$. Thus,
$$|AP|+|QB| \leq R+d$$
Moving $|AP|$ to the other side
$$|QB| \leq R + d - |AP| = d - (|AP| - R)$$
Now, $|AP| > R$ implies $|AP|-R>0$, so the right-hand side is strictly less than $d$:
$$|QB| \leq d - (|AP| - R) < d$$
which means
$$|QB| < d$$