已知△ABC和△A’B’C’中,AB=A'B',BC=B'C',中线为AD和A'D',AD=A'D'求证△ABC≌△A'B'C'
证明:延长AD和A'D'到E和E',使AD=DE,A'D'=D'E',连接BE B'E'
又∠ADC=∠BDE ∠A'D'C'=∠B'D'E',BD=CD,B'D'=C'D' ∴△A'D'C'≌△B'D'E' △ADC≌△BDE
∴AC=BE.A'C'=B'E' ∠C=∠CBE,∠C'=∠C'B'E' ∴BE=B'E'
AB=A'B',AE=A'E' ∴△ABE≌△A'B'E'∴∠ABE=∠ABC+∠CBE=∠ABC+∠C=180-∠BAC
同理∠A'B'E'=180-∠B'A'C',∵∠ABE=∠A'B'E',∴∠BAC=∠B'A'C',又∵AB=A'B',AC=A'C',
∴△ABC≌△A'B'C'(SAS)