$\because$点$P$是直线$AB$上一动点，
$\therefore$有以下三种情况：
①当点$P$在线段$AB$上时，如图$1$所示：

$\therefore \angle PDC=180^{\circ}-\angle 1$，$\angle PEC=180^{\circ}-\angle 2$，
$\because \angle C+\angle PDC+\angle PEC+\angle DPE=360^{\circ}$，$\angle C=90^{\circ}$，
$\therefore 90^{\circ}+180^{\circ}-\angle 1+180^{\circ}-\angle 2+\angle \alpha =360^{\circ}$，
$\therefore \angle 1+\angle 2-\angle \alpha =90^{\circ}$；
②当点$P$在$BA$的延长线上时，如图$2$所示：

$\because \angle PAD=\angle C+\angle ABC=90^{\circ}+\angle ABC$，$\angle APE=180^{\circ}-\angle 2-\angle ABC$，
$\therefore \angle APD=\angle DPE+\angle APE=\angle \alpha +180^{\circ}-\angle 2-\angle ABC$，
$\because \angle PAD+\angle APD+\angle PDA=180^{\circ}$，
$\therefore 90^{\circ}+\angle ABC+\angle \alpha +180^{\circ}-\angle 2-\angle ABC+\angle 1=180^{\circ}$，
$\therefore \angle 2-\angle 1-\angle \alpha =90^{\circ}$；
③当点$P$在$AB$的延长线上时，如图$3$所示：

$\because \angle PBE=\angle C+\angle BAC=90^{\circ}+\angle BAC$，$\angle BPD=180^{\circ}-\angle 1-\angle BAC$，
$\therefore \angle BPE=\angle DPE+\angle BPD=\angle \alpha +180^{\circ}-\angle 1-\angle BAC$，
$\because \angle BPE+\angle PEB+\angle PBE=180^{\circ}$，
$\therefore \angle \alpha +180^{\circ}-\angle 1-\angle BAC+\angle 2+90^{\circ}+\angle BAC=180^{\circ}$，
$\therefore \angle 1-\angle 2-\angle \alpha =90^{\circ}$，
综上所述：$\angle \alpha$，$\angle 1$，$\angle 2$之间的关系为：$\angle 1+\angle 2-\angle \alpha =90^{\circ}$或$\angle 2-\angle 1-\angle \alpha =90^{\circ}$或$\angle 1-\angle 2-\angle \alpha =90^{\circ}$.
故答案为：$\angle 1+\angle 2-\angle \alpha =90^{\circ}$或$\angle 2-\angle 1-\angle \alpha =90^{\circ}$或$\angle 1-\angle 2-\angle \alpha =90^{\circ}$.

$\because$点$P$是直线$AB$上一动点，
$\therefore$有以下三种情况：
①当点$P$在线段$AB$上时，如图$1$所示：

$\therefore \angle PDC=180^{\circ}-\angle 1$，$\angle PEC=180^{\circ}-\angle 2$，
$\because \angle C+\angle PDC+\angle PEC+\angle DPE=360^{\circ}$，$\angle C=90^{\circ}$，
$\therefore 90^{\circ}+180^{\circ}-\angle 1+180^{\circ}-\angle 2+\angle \alpha =360^{\circ}$，
$\therefore \angle 1+\angle 2-\angle \alpha =90^{\circ}$；
②当点$P$在$BA$的延长线上时，如图$2$所示：

$\because \angle PAD=\angle C+\angle ABC=90^{\circ}+\angle ABC$，$\angle APE=180^{\circ}-\angle 2-\angle ABC$，
$\therefore \angle APD=\angle DPE+\angle APE=\angle \alpha +180^{\circ}-\angle 2-\angle ABC$，
$\because \angle PAD+\angle APD+\angle PDA=180^{\circ}$，
$\therefore 90^{\circ}+\angle ABC+\angle \alpha +180^{\circ}-\angle 2-\angle ABC+\angle 1=180^{\circ}$，
$\therefore \angle 2-\angle 1-\angle \alpha =90^{\circ}$；
③当点$P$在$AB$的延长线上时，如图$3$所示：

$\because \angle PBE=\angle C+\angle BAC=90^{\circ}+\angle BAC$，$\angle BPD=180^{\circ}-\angle 1-\angle BAC$，
$\therefore \angle BPE=\angle DPE+\angle BPD=\angle \alpha +180^{\circ}-\angle 1-\angle BAC$，
$\because \angle BPE+\angle PEB+\angle PBE=180^{\circ}$，
$\therefore \angle \alpha +180^{\circ}-\angle 1-\angle BAC+\angle 2+90^{\circ}+\angle BAC=180^{\circ}$，
$\therefore \angle 1-\angle 2-\angle \alpha =90^{\circ}$，
综上所述：$\angle \alpha$，$\angle 1$，$\angle 2$之间的关系为：$\angle 1+\angle 2-\angle \alpha =90^{\circ}$或$\angle 2-\angle 1-\angle \alpha =90^{\circ}$或$\angle 1-\angle 2-\angle \alpha =90^{\circ}$.
故答案为：$\angle 1+\angle 2-\angle \alpha =90^{\circ}$或$\angle 2-\angle 1-\angle \alpha =90^{\circ}$或$\angle 1-\angle 2-\angle \alpha =90^{\circ}$.