$證明：?(1)?連接?DE、??DF，?如圖①$
$當(dāng)?t=2?時(shí)，?DH=AH=4，?則?H?是?AD?的中點(diǎn)$
$∵?EF⊥AD?$
$∴?EF ?為?AD?的垂直平分線$
$∴?AE= DE，??AF= DF?$
$∵?AB= AC，??AD⊥BC，?∴?∠B=∠C?$
$∴?EF//BC?$
$∴?∠AEF=∠B，??∠AFE=∠C?$
$∴?∠AEF=∠AFE?$
$∴?AE= AF?$
$∴?AE=AF= DE= DF?$
$∴四邊形?AEDF ?為菱形$
$?(2)?如圖②由?(1)?知?EF//BC?$
$∴?△AEF∽△ABC?$
$∴?\frac {EF}{BC}=\frac {AH}{AD}，?即?\frac {EF}{10}=\frac {8-2t}{8}?$
$解得?EF= 10-\frac {5t}{2}?$
$?S_{△PEF}=\frac {1}{2}EF×DH=\frac {1}{2}(10-\frac {5}{2}t)×2t=-\frac {5}{2}t2+10t=-\frac {5}{2}(t-2)2+10?$
$∴當(dāng)?t= 2s ?時(shí)，?S_{△PEF} ?取最大值，最大值為?10，?此時(shí)?BP=3t=6(\ \mathrm {cm})?$
$?(3)?存在，理由如下：$
$①若點(diǎn)?E?為直角頂點(diǎn)，如圖③$
$此時(shí)?PE//AD，??PE= DH= 2t，??BP= 3t?$
$∵?PE//AD?$
$∴?\frac {PE}{AD}=\frac {BP}{BD}，?即?\frac {2t}{8}=\frac {2t}{5}?$
此比例式不成立，故此種情形不存在
$②若點(diǎn)?F ?為直角項(xiàng)點(diǎn)，如圖④$
$此時(shí)?PE//AD，??PF=DH= 2t，??BP= 3t，??CP= 10- 3t?$
$∵?PF//AD?$
$∴?\frac {PF}{AD}=\frac {CP}{CD}，?即?\frac {2t}{8}=\frac {10-3t}{5}?$
$解得?t=\frac {40}{17}?$
$③若點(diǎn)?P ?為直角頂點(diǎn)，如圖⑤$
$過(guò)點(diǎn)?E?作?EM⊥BC，?垂足為?M，?過(guò)點(diǎn)?F ?作?FN⊥BC，?垂足為?N?$
$則?EM = FN = DH= 2t，??EM//FN//AD?$
$∵?EM//AD?$
$∴?\frac {EM}{AD}=\frac {BM}{BD}，?即?\frac {2t}{8}=\frac {BM}{5}?$
$解得?BM=\frac {5}{4}t?$
$∴?PM=BP-BM=3t-\frac {5}{4}t=\frac {7}{4}t?$
$在?Rt△EMP ?中，由勾股定理，$
$得?PE2=EM2+PM2= (2t)2+(\frac {7}{4}t)2=\frac {113}{16}t2?$
$∵?FN//AD?$
$∴?\frac {FN}{AD}=\frac {CN}{CD}，?即?\frac {2t}{8}=\frac {CN}{5}?$
$解得?CN=\frac {5}{4}t?$
$∴?PN= BC- BP- CN= 10- 3t-\frac {5}{4}t= 10-\frac {17}{4}t?$
$在?Rt△FNP ?中，由勾股定理得?PF2=FN2+PN2= (2t)2+(10-\frac {17}{4}t)2=\frac {353}{16}t2-85t+100?$
$在?Rt△PEF ?中，由勾股定理，得?EF2=PE2+PF2?$
$即?(10-\frac {5}{2}t)2=(\frac {113}{16}t2)+(\frac {353}{16}t2-85t+100)?$
$化簡(jiǎn)得?\frac {233}{8}t2-35t=0?$
$解得?t= \frac {280}{233}?或?t=0 (?舍去)$
$∴?t=\frac {280}{233}?$
$綜上所述，當(dāng)?t=\frac {40}{17}s ?或?t=\frac {280}{233}s ?時(shí)，?△PEF ?為直角三角形$
