②证明：$\because EF⊥ BC，$
$\therefore ∠ EFC=90°，$$∠ C+∠ CEF=90°。$
$\because ∠ A=90°，$
$\therefore ∠ C+∠ ABC=90°，$
$\therefore ∠ CEF=∠ ABC。$
$\because ∠ AEF=180°-2α，$
$\therefore ∠ CEF=2α，$
$\therefore ∠ ABC=2α。$
$\because BD$是$△ ABC$的角平分线，
$\therefore ∠ ABD=\frac{1}{2}∠ ABC=α，$
$\therefore ∠ ABD=∠ M，$
$\therefore BD// ME。$
(2)解：$2∠ BNE=90°+∠ BAC。$
证明：设$∠ ABD=x，$$∠ AEM=y，$
$\because BD$平分$∠ ABC，$$EM$平分$∠ AEF，$
$\therefore ∠ ABC=2x，$$∠ AEF=2y。$
$\because ∠ ABD+∠ BAD=180°-∠ ADB，$
$∠ NED+∠ END=180°-∠ NDE，$
$∠ ADB=∠ NDE，$
$\therefore ∠ ABD+∠ BAD=∠ NED+∠ END，$
$\therefore x+∠ BAD=y+∠ END，$
$\therefore x-y=∠ END-∠ BAD，$
同理$∠ ABC+∠ BAC=∠ FEC+∠ EFC，$
$\therefore 2x+∠ BAC=2y+∠ EFC，$
$\therefore 2x-2y=∠ EFC-∠ BAC。$
$\because EF⊥ BC，$
$\therefore ∠ EFC=90°，$
$\therefore 2(x-y)=90°-∠ BAC，$
$\therefore 2(∠ END-∠ BAD)=90°-∠ BAC，$
即$2(∠ BNE-∠ BAC)=90°-∠ BAC，$
$\therefore 2∠ BNE=90°+∠ BAC。$
