初中数学题,高手进

2024-12-21 23:12:19
推荐回答(5个)
回答1:

(1):因为∠ACB=90°
所以∠ACB=∠ACE
又因为CE=CD
所以△BDC全等△AEC (SAS)
所以AE=BD
(2):PS:我估计你是把D打成B了,B、E为AC边的三等分点没法做,应该是D、E三等分AC
解题如下:
因为D、E三等分AC
所以AD=CE=三分之一AC
因为DE=AC-AD-CE
所以DE=三分之一AC
即AD=DE
因为EF∥AB
所以∠BAE=∠AEF
又因为∠ADB=∠FDE (对顶角相等)
所以△ABD全等△FDE(ASA)
所以BD=DF
即点D是BF的中点

回答2:

第一题!由已知可得三角形ABC是一个等腰直角三角形,AC=BC,且CE=CD,所以三角形AEC和三角形BDC全等,所以AE与BD相等。
第二题!题面不对,无法解答!

回答3:

题一,相等,利用勾股定理,得AE2=AC2+EC2,BD2=DC2+BC2(这里的2指平方),又因为AC=BC,EC=DC,所以,AE=BD
题二,因为AB//EF,所以∠ABD=∠DFE,又因为∠ADB=∠EDF,AD=DE,所以△ABDA≌△DEF,所以BD=DF.D是BF的中点

回答4:

题1:在△ACE和△BCD中
AC=BC ∠ACE=∠BCD CE=CD
∴△ACE全等于△BCD(SAS)
∴AE=BD(全等三角形对应边相等)
题2:∵EF//AB∴∠BAD=∠FED
在△ABD和△EFD中
∠BAD=∠FED AD=ED ∠ADB=∠EDF
∴△ABD全等于△EFD(ASA)
∴BD=ED(去昂等三角形对应边相等)∴点D是BF的中点

回答5:

解:1、∵∠ACB=90°
∴∠ACE=90°
在△BCD与△ACE中
∵AC=BC、∠ACB=∠ACE=90°、CE=CD
∴△BCD≌△ACE(SAS)
∴AE=BD
2、∵B(D)、E为AC边的三等分点(题目打错了,B要改为D)
∴AD=DE=CE
∵∠ADB与∠EDF为对顶角
∴∠ADB=∠EDF
∵EF∥AB
∴∠BAD=∠DEF
在△ADE与△DEF中
∵∠ADB=∠EDF、∠BAD=∠DEF、AD=DE
∴△ADE≌△DEF(AAS)
∴BD=FD
∴点D是BF的中点。

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