$解：(3)CP= AM$
$证明：∵BE⊥AP，CF⊥AP，∴∠BEA=∠AFC$
$∴∠BAE+∠CAP=90°，∠ACF+∠CAP=90°$
$∴∠BAE=∠ACF$
$∵AB=AC，∴△BAE≌△ACF，∴AE=CF$
$又易知∠BAD=∠ACD=45°，∴∠EAM=∠FCP$
$在△CFP 和△AEM中$
$\begin{cases}{∠FCP=∠EAM }\\{CF=AE} \\ {∠CFP=∠AEM} \end{cases}$
$∴△CFP≌△AEM，∴CP=AM$
$(4)S_{△ABC}=\frac{1}{2}×BC×AD=4$
$由图形可知，S_{△ABC}= S_{△APB} + S_{△APC}=\frac{1}{2}×AP×BE+\frac{1}{2}×AP×CF=\frac 12×AP×(d_{1}+d_{2})$
$当AP⊥BC时，AP 最小，此时AP=2，∴d_{1}+d_{2}的最大值为\frac{2×4}{2}=4$
