(1)证明:∵EP平分∠BEC, ∴∠BEP=∠CEP.△ACE中,∠A+∠ACE+∠AEC=180°, ∵∠ACE=∠ACB+∠BCE,且∠A=∠ACB, ∴2∠A+2∠BEP+∠BCE=180°, ∴2(∠A+∠BEP)+∠BCE=180°, ∵∠CPD=∠A+∠BEP, ∴2∠CPD+∠BCE=180°, ∴∠CPD=90°- ∠BCE; (2)结论:∠CPD= ∠BCE, 理由如下:解:设∠CAB=∠ACB=α, ∵ED平分∠BEC,∴∠BED=∠CED, 设∠BED=∠CED=β, 则∠CEB=2β, 分两种情况:i)若点E在BA上(E不与A、B重合, 如图,∵∠ACE=∠ACB-∠BCE, ∴∠ACE=α-(2α-2β)=2β-α, ∴∠BCE=∠ACB-∠ACE=α-(2β-α)=2α-2β, ∵∠CPD=∠CED-∠ACE, ∴∠CPD=β-(2β-α)=α-β, ∴∠CPD= ∠BCE; ii)若E在BA的延长线上, 如图,∵∠ACE=∠CAB-∠CEB, ∴∠ACE=α-2β, ∴∠BCE=∠ACB+∠ACE=α+(α-2β)=2α-2β, ∵∠CPD=∠ACE+∠CEP, ∴∠CPD=α-2β+β=α-β, ∴∠CPD= ∠BCE, 综上,可知∠CPD= ∠BCE。 | ![](http://img.shitiku.com.cn/uploads/allimg/20191026/20191026112740-60195.png)
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