略

设 $A(x_{1},y_{1})$，则$$y_{1}=\dfrac{1}{2}x_{1}^{2}.$$由 $y=\dfrac{1}{2}x^{2}$ 得 $y'=x$，所以 $y'|_{x=x_{1}}=x_{1}$．
因此抛物线 $C$ 在点 $A$ 点处的切线方程为$$y-y_{1}=x_{1}(x-x_{1}),$$即$$y=x_{1}x-y_{1}.$$设 $P(x_{0},kx_{0}-1)$，则有$$kx_{0}-1=x_{0}x_{1}-y_{1}.$$设 $B(x_{2},y_{2})$，同理有$$kx_{0}-1=x_{0}x_{2}-y_{2},$$所以 $AB$ 的方程为$$kx_{0}-1=x_{0}x-y,$$即$$x_{0}(x-k)-(y-1)=0,$$所以直线 $AB$ 恒过定点 $Q(k,1)$．

略

$PQ$ 的方程为$$y=\dfrac{kx_{0}-2}{x_{0}-k}(x-k)+1,$$与抛物线方程 $y=\dfrac{1}{2}x^{2}$ 联立，消去 $y$，得\[x^{2}-\dfrac{2kx_{0}-4}{x_{0}-k}x+\dfrac{(2k^{2}-2)x_{0}-2k}{x_{0}-k}=0.\]设 $M(x_{3},y_{3})$，$N(x_{4},y_{4})$，则\[x_{3}+x_{4}=\dfrac{2kx_{0}-4}{x_{0}-k},x_{3}x_{4}=\dfrac{(2k^{2}-2)x_{0}-2k}{x_{0}-k}\cdots\cdots \text{ ① }\]需证$$|PM|\cdot |QN|=|PN|\cdot |QM|,$$即证$$\dfrac{PM}{PN}=\dfrac{QM}{QN},$$则只需证明\[\dfrac{x_{3}-x_{0}}{x_{4}-x_{0}}=\dfrac{k-x_{3}}{x_{4}-k},\]即\[2x_{3}x_{4}-(k+x_{0})(x_{3}+x_{4})+2kx_{0}=0\cdots\cdots\text{ ② }\]由 ① 知，\[\begin{split}&\quad 2x_{3}x_{4}-(k+x_{0})(x_{3}+x_{4})+2kx_{0}\\&=\dfrac{2(2k^{2}-2)x_{0}-4k}{x_{0}-k}-(k+x_{0})\dfrac{2kx_{0}-4}{x_{0}-4}+2kx_{0}\\&=\dfrac{2(2k^{2}-2)x_{0}-4k-(k+x_{0})(2kx_{0}-4)+2kx_{0}(x_{0}-k)}{x_{0}-k}\\&=0.\end{split}\]故 ② 式成立，从而结论成立．