证明：$(1)$∵$CE$是$△ABC$的角平分线
∴$∠ECB=∠ACE$
又∵$∠AEF = ∠B + ∠ECB，$$∠AFE = ∠FAC + ∠ACE，$$ ∠B=∠FAC$
∴$ ∠AEF=∠AFE$
$(2)∠P+ ∠CFD =90°$
∵$CP $是$∠ACQ $的平分线
∴$ ∠ACP=\frac {1}{2}∠ACQ$
∵$∠ACE = \frac {1}{2}∠ACB$
∴$ ∠ECP = ∠ACE + ∠ACP =\frac {1}{2}(∠ACB+∠ACQ)=90°$
∴$∠P+∠AEC=90°$
∵$∠AEF= ∠AFE = ∠CFD$
∴$∠P + ∠CFD = 90°$
$(3)$延长$PE$交$BC$于点$H，$设$PE$交$AC$于点$K$
∵$PE$平分$∠BPD$
∴$ ∠BPK =∠KPF$
又∵$ ∠EKC = ∠KPF +∠PFA，$$∠EHC =∠B+∠BPK，$$∠B= ∠CFD=∠PFA$
∴$∠EKC= ∠EHC$
∵$CE⊥KH$
∴$ ∠CEK = ∠CEH = 90°$
∴$ ∠EKC + ∠ECK = 90°，$$∠EHC + ∠ECH = 90° $
∴$∠ECK=∠ECH$
∴$CE$平分$∠ACB$
