如图,若E为BA延长线上一动点,连EC∠AEC与∠ACE的角平分线交于Q，当E滑动时有下面两个结论

如图,若E为BA延长线上一动点,连EC∠AEC与∠ACE的角平分线交于Q，当E滑动时有下面两个结论

由题意知
Q点是△AEC的内心（内角平分线的交点）
连接AQ AQ平分∠CAE
∠CQE=∠CAE+1/2(∠AEC+∠ACE)
=1/2(∠CAE+∠AEC+∠ACE)+1/2∠CAE
=90+1/2∠CAE
又∠CAE=∠ABC+∠ACB
∠CQE=90+1/2(∠ABC+∠ACB)
∵BA1是∠CBA的平分线
∠DCA1=∠A1+∠A1BC
=∠A1+1/2∠CBA
又CA1是∠ACD的平分线
∠DCA1=1/2∠ACD=1/2(∠CBA+∠BAC)
∴∠A1+1/2∠CBA=1/2∠CBA+1/2∠BAC
∠A1=1/2∠BAC

∠CQE+∠A1=90+1/2∠CBA+1/2∠BCA+1/2∠BAC
=90+1/2(∠CBA+∠BCA+∠BAC)
=90+90
=180
是一个定值

结论1正确