解?$：(2)$?方法一?$：$?設?$△ABD$?的高為?$h_{1}$?
?$S_{△ACE}$?的高為?$h_{2}$?
?$S_{△ABD}+S_{△ACE}$?
?$=\frac {1}{2}×BD×h_{1}+\frac {1}{2}×CE×h_{2}$?
?$=\frac {1}{2}×BD×(h_{1}+h_{2})$?
?$=\frac {1}{2}×BD×DE$?
設?$△ADE$?的高為?$h_{3}，△ABC$?的高為?$h_{4}$?
∴?$S_{△ADE}-S_{△ABD}-S_{△ACE}$?
?$=\frac {1}{2}×DE×h_{3}-\frac {1}{2}×BD×DE$?
?$=\frac {1}{2}×DE×(h_{3}-BD)$?
?$=\frac {1}{2}×DE×h_{4}$?
?$=\frac {1}{2}×BC×h_{4}$?
?$=S_{△ABC}$?
方法二?$：S_{△ABD}+S_{△ACE}$?
?$=S_{五邊形ABDEC}-S_{△ADE}$?
?$=S_{△ABC}+S_{四邊形BDEC}-S_{△ADE}$?
?$S_{△ADE}-S_{△ABD}-S_{△ACE}$?
?$=S_{△ADE}-S_{△ABC}-S_{四邊形BDEC}+S_{△ADE}$?
?$=2S_{△ADE}-S_{△ABC}-S_{四邊形BDEC}$?
?$=2×\frac {1}{2}×DE×(BD+h_{4})-S_{△ABC}-S_{四邊形BDEC}$?
?$=DE×BD+DE×h_{4}-S_{△ABC}-S_{四邊形BDEC}$?
?$=S_{四邊形BDEC}+2△ABC-S_{△ABC}-S_{四邊形BDEC}$?
?$=S_{△ABC}$?
