解：$(2)∠E+∠F=180°$
理由如下：∵$∠BAD+∠CDA+∠ABC+∠BCD=360°，$$∠BAD，$
$∠CDA$的平分线交于点$E，$$∠ABC，$$∠BCD$的平分线交于点$F$
∴$∠DAE+∠ADE+∠FBC+∠BCF= \frac {1}{2} (∠BAD+∠CDA +$
$∠ABC+∠BCD)=180°$
∵$∠DAE+∠ADE+∠E=180°，$$∠FBC+∠BCF+∠F=180°$
∴$∠DAE+∠ADE+∠E+∠FBC+∠BCF+∠F=360°$
∴$∠E+∠F=360°-(∠DAE+∠ADE+∠FBC+∠BCF)=180°$
