Question

問題探究】

（1）如圖1，銳角△ABC中，分別以AB、AC為邊向外作等腰△ABE和等腰△ACD，使AE=AB，AD=AC，∠BAE=∠CAD，連接BD，CE，試猜想BD與CE的大小關(guān)系，并說明理由．

【深入探究】

（2）如圖2，四邊形ABCD中，AB=7cm，BC=3cm，∠ABC=∠ACD=∠ADC=45º，求BD的長．

（3）如圖3，在(2)的條件下，當(dāng)△ACD在線段AC的左側(cè)時(shí)，求BD的長．

 

 

Answer 1

（1）答：BD =CE． ················································································································· 1分

理由：∵∠BAE=∠CAD，

∴∠BAE+∠BAC=∠CAD+∠BAC，即∠EAC=∠BAD，··························································· 2分

又∵AE=AB，AC=AD，

∴△EAC≌△BAD  (SAS) ，

∴BD=CE． ··························································································································· 4分

（2）解：如圖1，在△ABC的外部，以點(diǎn)A為直角頂點(diǎn)作等腰直角三角形BAE，使∠BAE=90º，AE=AB，連接EA、EB、EC． ····································································································································· 5分

∵，

∴，，

∴∠BAE=，

∴∠BAE+∠BAC=∠CAD+∠BAC，

即∠EAC=∠BAD，

∴△EAC≌△BAD  (SAS) ， ·························· 7分

∴BD=CE．

∵AE=AB=7，

∴， ∠AEC=∠AEB=45º．

又∵∠ABC=45º，

∴∠ABC+∠ABE=45º+45º=90º， ···························································································· 8分

∴EC==，

∴．

答：BD長是cm． ········································································································ 9分

(3)如圖2，在線段AC的右側(cè)過點(diǎn)A作AE⊥AB于A，交BC的延長線于點(diǎn)E， ···················· 10分

∴∠BAE=90º，

又∵∠ABC=45º，

∴∠E=∠ABC=45º，

∴AE=AB=7，．····················································································· 11分

又∵∠ACD=∠ADC=45 º，

∴∠BAE= ∠DAC=90º，

∴∠BAE∠BAC=∠DAC∠BAC，

即∠EAC=∠BAD，

∴△EAC≌△BAD  (SAS) ，

∴BD=CE． ····································· 13分

∵BC=3，

∴BD=CE=(cm)．

BD長是()cm．