有一條求證題

2011-01-10 5:32 am

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AEB 和BDC 係直綫,AC=BC,AD 係角BAC 既角平分綫,DE 垂直AB 及AC 垂直BC A. 求證 DC =DE B 若AB 係8CM ,求三角形BDE 周界

更新1:

樓下唔係aas 咩?

回答 (5)

用戶2053

2011-01-10 6:03 am

✔ 最佳答案

A)AD = AD (公共)
ㄥDAE = ㄥDAC (已知)
ㄥDEA = ㄥDCA = 90° (已知)△DAE ~= △DCA (A.S.A)對應邊 DC =DE 。
B)△BDE 周界= EB + BD + DE
= EB + BD + DC
= EB + BC
= (AB + BC + CA) - (EA + AC) , △DAE ~= △DCA 故 EA = AC ,
= (8 + 2BC) - (2AC)
= (8 + 2BC) - (2BC)
= 8

2011-01-09 22:06:47 補充：
三角形BDE 周界
= 8CM

2011-01-09 22:12:17 補充：
B)

另法 :
△BDE 周界
= EB + BD + DE
= EB + BD + DC
= EB + BC
= EB + AC , △DAE ~= △DCA 故 EA = AC ,
= EB + EA
= AB
= 8 CM

2011-01-09 22:51:57 補充：
aas , sas 好像不一定要順序?
sorry不太肯定。

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