$证明: (1)∵BE,CF 是△ABC的高$
$∴∠AEB=90°,∠AFC=90°$
$∴∠ABP+∠BAE=90°,∠ACQ+∠BAE=90°$
$∴∠ABP=∠ACQ$
$在△ABP 与△QCA中$
$\begin{cases}{BP=CA }\\{∠ABP=∠QCA} \\ {AB=QC} \end{cases}$
$∴△ABP≌△QCA(\mathrm {SAS})$
$(2)AP⊥AQ$
$由△ABP≌△QCA得∠BAP= ∠Q$
$∵∠Q+∠BAQ=90°$
$∴∠BAP+∠BAQ =90°,即∠PAQ = 90°$
$∴PA⊥AQ$