每日一專題

【例1】∴△ABC為等邊三角形,∴∠B=∠C=∠DEF=60°,∵∠CED=∠DEF+∠CEF=∠B+∠BDE,∴CEF= BDE,.'. DBE ECF(ASA), . BE=CF.
【例2】易知∠CDE=∠A=∠B=45°,∠BDE=∠ACD,CD=DE, . AACDABDE(AAS), .' BD=AC=BC.
1.由∠AFE=∠ABC=∠C=60°,可得∠CBE=∠BAD,
AB= BC,.'. ABD BCE, .' AD= BE
2.過(guò)點(diǎn)A作AE⊥CD于點(diǎn)E,可得△ACE≌△CBD,∴AE=CD=4,∴.S△Ac=2CD·AE=2×4×4=8.
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