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$\because GP\perp PQ$
$\therefore \angle APG=\angle GPM=90\degree$
$\text{又}\because \angle 1+\angle 2=\angle APG, \angle 3+\angle SPQ=\angle GPM$
$\therefore \angle 1=\angle SPQ$
$\text{又}\because AB, S'P\text{交于点}P$
$\therefore \angle 1=\angle S'PQ$
$\therefore \angle SPQ=\angle S'PQ$
$\text{同理，}\angle SQP=\angle S'QP$

$\because \text{在}\triangle SPQ\text{和}\triangle S'PQ\text{中}$
$\begin{cases} \angle SPQ=\angle S'PQ \\ PQ=PQ \\ \angle SQP=\angle S'QP\end{cases}$
$\triangle SPQ≌\triangle S'PQ(ASA)$
$\therefore SP=S'P$

$\because \text{在}\triangle SPM\text{和}\triangle S'PM\text{中}$
$\begin{cases} SP=S'P \\ \angle SPM=\angle S'PM \\ PM=PM\end{cases}$
$\triangle SPM≌\triangle S'PM(SAS)$
$\therefore SM=S'M, \angle SMP=\angle S'MP$

$\text{又}\because M\text{在}SS'\text{上}$
$\therefore \angle SMP+\angle S'MP=180\degree$
$\therefore \angle SMP=\angle S'MP=180\degree \div 2=90\degree$
$\therefore AB \perp SS'$