如图,已知AD=AB,AC=AE,角DAB=角EAC=90度,连结DC、BE.请说出线段DC与BE的位置关系
15如图,AB=AC,AD=AE.AB、DC相交于M,AC、BE相交于N,∠DAB=∠EAC.求证:AM=AN
如图,已知AB=AC,AD=AE,AB与DC相交于点M,∠DAB=∠EAC,AM=5,求AN的长
如图,在△ABC外有△ABD和△ACE,且∠DAB=∠EAC=90°,AD=AB,AC=AE,DC与BE交于M,求,MA
已知:如图AB=AC,AD=AE.AB,DC相交于点M,AC,BE相交于点N,∠DAB=∠EAC.求证:∠D=∠E
如图,AB=AC,AD=AE,CD=BE,求证,角DAB=角EAC
如图6,AB=AC,AD=AE,∠DAB=∠EAC,AB、DC相交于点M,AC、BE相交于点N,BE、CD相交于点O,证
如图,AB=AC,AD=AE,AB,DC相交于点M,AC,BE相交于点N,∠DAB=∠EAC.求证:△ADM≌△AEN.
真么写已知如图ab=ac ad=ae ab dc相交于点m ac be相交于点n,角bab=角eac 求角am=an
已知如图,AB=AC,AD=AE,AB、DC相交于点M,AC、BE相交于点N,AP⊥DC于P,AQ⊥BE于Q且∠DAB=
已知,AB=AD,AC=AE,角1=角2,求证DC=BE
如图,已知AB=AD,AC=AE,∠DAB=∠EAC,BE,CD交于P,求证:AP平分∠DPE
如图,在四边形ABCD中,AB=DC,延长线段CB到E,是BE=AD,连接AE、AC,AE=AC,求证:AD‖EC