解$：(2)$方法一$：$设$△ABD$的高为$h_{1}$
$S_{△ACE}$的高为$h_{2}$
$S_{△ABD}+S_{△ACE}$
$=\frac {1}{2}×BD×h_{1}+\frac {1}{2}×CE×h_{2}$
$=\frac {1}{2}×BD×(h_{1}+h_{2})$
$=\frac {1}{2}×BD×DE$
设$△ADE$的高为$h_{3}，△ABC$的高为$h_{4}$
∴$S_{△ADE}-S_{△ABD}-S_{△ACE}$
$=\frac {1}{2}×DE×h_{3}-\frac {1}{2}×BD×DE$
$=\frac {1}{2}×DE×(h_{3}-BD)$
$=\frac {1}{2}×DE×h_{4}$
$=\frac {1}{2}×BC×h_{4}$
$=S_{△ABC}$
方法二$：S_{△ABD}+S_{△ACE}$
$=S_{五边形ABDEC}-S_{△ADE}$
$=S_{△ABC}+S_{四边形BDEC}-S_{△ADE}$
$S_{△ADE}-S_{△ABD}-S_{△ACE}$
$=S_{△ADE}-S_{△ABC}-S_{四边形BDEC}+S_{△ADE}$
$=2S_{△ADE}-S_{△ABC}-S_{四边形BDEC}$
$=2×\frac {1}{2}×DE×(BD+h_{4})-S_{△ABC}-S_{四边形BDEC}$
$=DE×BD+DE×h_{4}-S_{△ABC}-S_{四边形BDEC}$
$=S_{四边形BDEC}+2△ABC-S_{△ABC}-S_{四边形BDEC}$
$=S_{△ABC}$
