教师讲解错误
错误详细描述：
如图，在△ABC中，AB＝AC，点P是BC边上任意一点，且PE∥AC，PF∥AB，PE、PF分别交AB、AC于点E、F，你能说明PE、PF和AB之间有什么关系吗？
【思路分析】
推出平行四边形PEAF，推出PF=AE，∠EPB=∠C，根据等腰三角形的判定和性质推出PE=BE即可。
【解析过程】
结论是PE+PF=AB，证明：如图所示∵PE∥AC，PF∥AB，∴四边形PEAF是平行四边形，∴PF=AE，∠EPB=∠C，∵AC=AB，∴∠B=∠C，∴∠EPB=∠B，∴PE=BE，∵BE+AE=AB，∴PE+PF=AB．
结论是PE+PF=AB，证明：如图所示∵PE∥AC，PF∥AB，∴四边形PEAF是平行四边形，∴PF=AE，∠EPB=∠C，∵AC=AB，∴∠B=∠C，∴∠EPB=∠B，∴PE=BE，∵BE+AE=AB，∴PE+PF=AB．
考查了平行四边形的性质和判定和等腰三角形的性质和判定，证此题的关键是证PE=BE和PF=FC。
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京ICP备号 京公网安备如图,在三角形abc中,ab=ac,p是bc边上一点,PE丄AB于E,PF丄AC于F,BD是AC边上的高.试探究PE＋PF与BD之间的数量关系._百度作业帮
拍照搜题，秒出答案
如图,在三角形abc中,ab=ac,p是bc边上一点,PE丄AB于E,PF丄AC于F,BD是AC边上的高.试探究PE＋PF与BD之间的数量关系.
如图,在三角形abc中,ab=ac,p是bc边上一点,PE丄AB于E,PF丄AC于F,BD是AC边上的高.试探究PE＋PF与BD之间的数量关系.
连接PA∵PE⊥AB,PF⊥AC,BD⊥ACAB=AC∴S△ABP+S△ACP=S△ABC1/2AB×PE+1/2AC×PF=1/2BD×AC∴PE+PF=BD
法一：过P作PO⊥BD与O易证四边形OBFP为矩形则OB=PF∵AB=AC∴∠ABC=∠C∵PE⊥AB,PF⊥AC,∴∠EPB+∠ABC=90°，∠FPC+∠C=90°∴∠EPB=∠FPC则Rt△BEPQ≌Rt△BOP∴BE=BO∴BD=BO+OD=PF+PE法二：延长...已知A、B是圆x2+y2=1与x轴的两个交点，CD是垂直于AB的动弦，直线AC和DB相交于点P，问是否存在两个定点E、F，使||PE|-|PF||为定值？若存在，求出E、F的坐标； 若不存在，请说明理由．_百度作业帮
拍照搜题，秒出答案
已知A、B是圆x2+y2=1与x轴的两个交点，CD是垂直于AB的动弦，直线AC和DB相交于点P，问是否存在两个定点E、F，使||PE|-|PF||为定值？若存在，求出E、F的坐标； 若不存在，请说明理由．
已知A、B是圆x2+y2=1与x轴的两个交点，CD是垂直于AB的动弦，直线AC和DB相交于点P，问是否存在两个定点E、F，使||PE|-|PF||为定值？若存在，求出E、F的坐标；&若不存在，请说明理由．
由已知得A（-1，0）、B（1，0），设P（x，y），C（x0，y0），则D（x0，-y0），由A、C、P三点共线得0x0+1&&&&&&&&&&&&&&&&&①…（2分）由D、B、P三点共线得0x0-1&&&&&&&&&&&&&&&&②…（4分）①×②得2x2-1＝-y20x20-1&&&&&&&&&&&&&&&&&&&&&&&&&&&③又&x02+y02=1，∴y02=1-x02&&&代入③得&&x2-y2=1，即点P在双曲线x2-y2=1上，…（10分）故由双曲线定义知，存在两个定点E（-，0）、F（，0）（即此双曲线的焦点），使||PE|-|PF||=2（即此双曲线的实轴长）&为定值．　…（13分）
本题考点：
三点共线；双曲线的定义．
问题解析：
设P（x，y），C（x0，y0），则D（x0，-y0），由A、C、P三点共线得直线方程，由D、B、P三点共线得直线方程，将两式相乘得2x2-1＝-y20x20-1&而&x02+y02=1，从而y02=1-x02&即点P在双曲线x2-y2=1上，根据双曲线定义知，存在两个定点E（-，0），F（，0）满足条件．Always within reach
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PadFone mini accessory: PadFone mini Case/ Bumper Case/ TriCover如图,已知三角形abc中,ab=ac,p是bc上一点,pe垂直ab于点e,pf垂直ac于点f,cg垂直ab于点g.求证pe+pf=c_百度作业帮
拍照搜题，秒出答案
如图,已知三角形abc中,ab=ac,p是bc上一点,pe垂直ab于点e,pf垂直ac于点f,cg垂直ab于点g.求证pe+pf=c
如图,已知三角形abc中,ab=ac,p是bc上一点,pe垂直ab于点e,pf垂直ac于点f,cg垂直ab于点g.求证pe+pf=c
证明：连接AP∵PE⊥AB∴S△ABP＝AB×PE/2∵PF⊥AC,AB＝AC∴S△ACP＝AC×PF/2＝AB×PF/2∵CG⊥AB∴S△ABC＝AB×CG/2∵S△ABP+ S△ACP＝S△ABC∴AB×PE/2+ AB×PF/2＝AB×CG/2∴PE+PF＝CG数学辅导团解答了你的提问,