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Các câu hỏi dưới đây có thể giống với câu hỏi trên
8 tháng 2 2016
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! ^_^
30 tháng 1 2021
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)
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xin lỗi mk ko bt giải vì chưa ok
ai như vậy thì k mk nha