(2)解:AF//DC,理由如下: ∵DA//BF ∴∠DAF+∠F=180° ∵△ADC≌△AFB ∴∠D=∠F ∴∠DAF+∠D=180° ∴AF//DC
$ 解:∵△ABC≌△ADE$ $∴∠CAB=∠EAD=\frac {1}{2}(∠BAE-∠DAC)=55°$ $∴∠DFB=∠DAB+∠B=∠DAC+∠CAB+∠B=90°$ $∴∠DGB=∠DFB-∠D=65°$
解:由题可得,△ABC≌△ADC≌△ABE ∴∠CBA=∠EBA,∠ACB=∠ACD 易知,∠ABC+∠ACB=180°-∠BAC=30° ∴∠1=∠EBC+∠DCB=2(∠ABC+∠ACB)=60°
解:∵△ABC≌△DEF ∴AB=DE,BC=EF=30 g,AC=DF=25 g ∴在△ABC中可得, BC-AC<AB<BC+AC 即 5 g<AB< 55g ∴5g<m<55g
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