21. 在三角形$ABE$中,$AE\perp BE$,直线$CD// AB$.
(1) 如图①,点$E在直线CD$上,若$∠BAE= 60^{\circ}$,求$∠BED$的度数.
(2) 如图②,点$E在直线CD$的下方,$EB交CD于点F$,$G是AB$上一点,连接$GE交CD于点H$,点$K在AB$,$CD之间且在GH$的右侧,连接$GK$,$FK$. 若$GE$,$FB分别是∠AGK和∠KFD$的平分线,试说明$∠GKF= 2∠AEG$.
(3) 在(1)的条件下,点$P$,$Q在直线CD$上,点$P在点Q$左侧,$∠PAQ= 80^{\circ}$,$AM平分∠PAE交CD于点M$,$N是直线AB$上方一点,$∠NAB= 2∠BAQ$. 若$∠NAM= 150^{\circ}$,请直接写出$∠AQC$的度数.

解：(1) $\angle BED = 30^{\circ}$.
(2) 如图，作 $EM // CD$，作 $KN // AB$，
又 $\because AB // CD$，
$\therefore EM // AB // CD // KN$.
设 $\angle AGK = 2x$，$\angle KFD = 2y$，
又 $\because GE$，$FB$ 分别平分 $\angle AGK$，$\angle KFD$，
$\therefore \angle AGE = \angle KGE = \frac{1}{2} \angle AGK = x$，
$\angle KFB = \angle DFB = \frac{1}{2} \angle KFD = y$.
$\because AB // EM$，
$\therefore \angle GEM = \angle AGE = x$.
$\because CD // EM$，
$\therefore \angle FEM = \angle DFB = y$.
$\therefore \angle GEF = \angle GEM - \angle FEM = x - y$.
又 $\because AE \perp BE$，
$\therefore \angle AEG = 90^{\circ} - \angle GEF = 90^{\circ} - (x - y) = 90^{\circ} - x + y$.
$\therefore 2 \angle AEG = 2(90^{\circ} - x + y) = 180^{\circ} - 2x + 2y$.
$\because KN // AB$，
$\therefore \angle GKN + \angle AGK = 180^{\circ}$.
$\therefore \angle GKN = 180^{\circ} - \angle AGK = 180^{\circ} - 2x$.
$\because KN // CD$，
$\therefore \angle NKF = \angle KFD = 2y$.
$\therefore \angle GKF = \angle GKN + \angle NKF = 180^{\circ} - 2x + 2y$，
$\therefore \angle GKF = 2 \angle AEG$.
(3) $\angle AQC = 32^{\circ}$.