cho tam giác abc cân tại a.trên cạnh ab lấy điểm d.trên tia đối của tia ca lấy e sao cho bd=ce.de cắt bc tại i trên tia đối của bc lấy k sao cho bk=ci a) chưngs minh dbk=eci.
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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xin lỗi mk ko bt giải vì chưa ok
ai như vậy thì k mk nha
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)