Question

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

Answer 1

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