Cho tam giác ABC cân tại A.Trên cạnh AB lấy điểm D.Trên tia đối của tia CA lấy điểm E sao cho BD=CE.DE cắt BC tại I.Trên tia đối của BC lấy K sao cho BK=CI. a)chứng minh tam giác DBK bằng tam giác ECI. b) chứng minh tam giác KDI cân tại D. c) Vẽ tia Bx vuông góc với AB tại B.Qua A vẽ đường thẳng vuông góc với BC cắt BX tại O.CMR tam giác OBD=tam giác OCE
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
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