26.已知,在Rt△ABC中,∠ABC=90°.
(1)如圖1,延長(zhǎng)AB至點(diǎn)D,連接CD,取CD的中點(diǎn)E,過(guò)點(diǎn)E作EF⊥CB,求證:CF=BF;
(2)如圖2,分別以AB,AC為邊向外作等邊△ABD和等邊△ACE,連接BE,連接DC交AB于點(diǎn)M,交BE于點(diǎn)F,連接AF,求證:DF=BF+AF;
(3)如圖3,在△ABC內(nèi)部有一點(diǎn)P,當(dāng)P滿足PA+PB+PC的和最小時(shí),直接寫(xiě)出∠APC的值.