G8 (2-7, German Democratic Republic) Given an isosceles trapezoid, its two bases and height are $a, b$ and $h$.
(a) On the axis of symmetry of the trapezoid, find a point $P$ such that the angle subtended by the point to the legs of the trapezoid is a right angle.
(b) Find the distances from point $P$ to the two bases of the trapezoid.
(c) Under what conditions can point $P$ be constructed (i.e., discuss when point $P$ exists).
Given the isosceles trapezoid with the upper base $AD$ and the lower base $BC$ being $a$ and $b$ respectively, and the height being $h$, $MN$ is the axis of symmetry of trapezoid $ABCD$.
If $\angle DPC = 90^{\circ}$, then $P$ must lie on the circle with $CD$ as its diameter.
(a) Now, construct a circle with $CD$ as its diameter, intersecting the axis $MN$ at two points ($P$ and $P'$), or at one point $P$, or having an intersection.
(b) It is easy to see that Rt $\triangle DMP \sim \text{Rt} \triangle PNC$,

Thus,
$$
\left\{\begin{array}{l}
PM \cdot PN = MD \cdot NC = \frac{ab}{4}, \\
PM + PN = h.
\end{array}\right.
$$

According to Vieta's formulas, $PM$ and $PN$ are the roots of the equation
$$
x^{2} - hx + \frac{ab}{4} = 0.
$$

And by Vieta's formulas, as long as the above quadratic equation has roots, they must be two positive roots. Therefore, $P$ exists only when $\Delta \geqslant 0$, i.e.,
$$
\begin{array}{c}
h^{2} - ab \geqslant 0, \\
x = \frac{h}{2} \pm \sqrt{h^{2} - ab}.
\end{array}
$$

One of the roots is $PM$, and the other is $PN$.
(c) From (b), we know that
if $h^{2} > ab$, there are two points $P, P'$ that satisfy the condition, i.e., the circle intersects the line segment $MN$. For example, as shown in the figure, we can assume
$$
\begin{array}{l}
PM = P'N = \frac{h}{2} - \sqrt{h^{2} - ab}, \\
PN = P'M = \frac{h}{2} + \sqrt{h^{2} - ab}.
\end{array}
$$