(1)证明:连接AP.
∵AB=AC,
∴S△ABC=S△ABP+S△ACP=12AB×PD+12AC×PE=12×AB×(PD+PE),
∵S△ABC=12AB×CF,
∴PD+PE=CF.
CF+PE=PD.
P点在BC的延长线上,过P做AB⊥PD,过C作AB⊥CF,过P作PE⊥AC,交AC的延长线于E点,连接AP
∵AB=AC,
∴S△APB=S△ABC+S△ACP=12AB×CF+12AC×PE=12×AB×(CF+PE),
∵S△APB=12AB×PD,
∴CF+PE=PD.
(1)证明:连接AP.
∵AB=AC,
∴S△ABC=S△ABP+S△ACP=12AB×PD+12AC×PE=12×AB×(PD+PE),
∵S△ABC=12AB×CF,
∴PD+PE=CF.
CF+PE=PD.
P点在BC的延长线上,过P做AB⊥PD,过C作AB⊥CF,过P作PE⊥AC,交AC的延长线于E点,连接AP
∵AB=AC,
∴S△APB=S△ABC+S△ACP=12AB×CF+12AC×PE=12×AB×(CF+PE),
∵S△APB=12AB×PD,
∴CF+PE=PD.