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Ta có: \(\widehat{ABK}+\widehat{ABC}=180^0\)(hai góc kề bù)

\(\widehat{ECB}+\widehat{ACB}=180^0\)(hai góc kề bù)

mà \(\widehat{ABC}=\widehat{ACB}\)(hai góc ở đáy của ΔABC cân tại A)

nên \(\widehat{ABK}=\widehat{ECB}\)

hay \(\widehat{DBK}=\widehat{ECI}\)(đpcm)

24 tháng 2 2019

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xin lỗi mk ko bt giải vì chưa ok 

ai như vậy thì k mk nha

a/ Ta có tam giác ABC cân tại A

=> góc ABC=góc ACB

Mà +góc ABC+góc ABF=180 độ

      + góc ACB+góc BCE=180 độ

=> góc DBF=góc BCE

Xét tam giác BFD và tam giác CIE có

BD=CE(gt)

góc DBF=góc ECI(chứng minh trên)

FB=CI(gt)

Vậy tam giác BFD=tam giác CIE(c-g-c)

Làm rồi nhưng mk chắc chắn! ^_^

1:

a: Xét ΔABC có AD là phân giác

nên BD/AB=CD/AC

mà AB<AC

nên BD<CD

b: AB<AC
=>góc B>góc C

góc ADB=góc C+góc CAD

góc ADC=góc B+góc BAD

mà góc C<góc B và góc CAD=góc BAD

nên góc ADB<góc ADC

6 tháng 2 2022

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