As shown in the figure, in triangle ABC and triangle def, Ag and DH are respectively high, and ab = De, Ag = DH, ∠ BAC = ∠ EDF
prove:
∵AG⊥BC,DH⊥EF
∴∠AGB=∠DHE=90
∵AB=DE,AG=DH
∴△ABG≌△DEH(HL)
∴∠B=∠E
∵∠BAC=∠EDF
∴△ABC≌△DEF(ASA)
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