∴∠P=∠DAC,∠PFA=∠DAF,
∵AD平分∠BAC,
∴∠DAC=∠DAF,
∴∠P=∠PFA,
∴AP=AF,
∴△APF是等腰三角形.
(2)△DCH≌△BEF.
证明:∵AB∥CH,
∴∠BAD=∠H(两直线平行,内错角相等),∠B=∠DCH(两直线平行,内错角相等),
又∵EF∥AD(已知),
∴∠BFE=∠BAD;
∴∠BFE=∠H,
∵EF∥AD,
∴∠BEF=∠BAD,
又∵∠BDA=∠CDH(对顶角相等),
∴∠BEF=∠CDH,
∴∠BEF=∠CDH
则在△DCH和△BEF中,
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∴△DCH≌△BEF.
(3)AB=PC,
理由:∵AD平分∠BAC,
∴∠BAD=∠HAC,
∵AB∥CH,
∴∠HAC=∠H,
∴AC=CH,
∴△BEF≌△CDH,
∴BF=CH,
∴AC=BF,
∵△APF为等腰三角形,
∴AP=AF,
∴AC+AP=BF+AF,即AB=PC.