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=∠EAC,说明AP=AQ
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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=∠EAC,说明AP=AQ已知如图,AB=AC,AD=AE,AB、DC相交于点M,
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=∠EAC,说明AP=AQ
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=∠EAC,说明AP=AQ
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=∠EAC,说明AP=AQ
证明:
∵∠DAC=∠BAC+∠DAB,∠EAB=∠BAC+∠EAC,∠DAB=∠EAC
∴∠DAC=∠EAB
∵AB=AC,AD=AE
∴△ABE≌△ACD (SAS)
∴BE=CD,S△ABE=S△ACD
∵AP⊥DC,AQ⊥BE
∴S△ABE=BE×AQ/2,S△ACD=CD×AP/2
∴BE×AQ/2=CD×AP/2
∴AP=AQ
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