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a) Ta có: ΔABC cân tại A(gt)
nên \(\widehat{ABC}=\widehat{ACB}\)(hai góc ở đáy)
mà \(\widehat{ACB}=\widehat{ECK}\)(hai góc đối đỉnh)
nên \(\widehat{ABC}=\widehat{ECK}\)
hay \(\widehat{DBH}=\widehat{ECK}\)
Xét ΔDBH vuông tại H và ΔECK vuông tại K có
BD=CE(gt)
\(\widehat{DBH}=\widehat{ECK}\)(cmt)
Do đó: ΔDBH=ΔECK(Cạnh huyền-góc nhọn)
Suy ra: BH=CK(Hai cạnh tương ứng)
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
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
Từ D kẻ DI // AE
Vì ΔABC cân tại A nên Bˆ = ACBˆ (1)
Vì DI // AE => ACBˆ = DIBˆ (đồng vị ) (2)
Từ (1) và (2) => Bˆ = DIBˆ
Trong ΔDIB có : Bˆ = DIBˆ => ΔDIB cân tại D
=> DB = DI mà DB = CE (gt)
=> DI = CE
Vì DI // AE => MDIˆ = MECˆ(so le trong )
và DIMˆ = MCEˆ ( so le trong )
Xét ΔDIM và ΔECM có :
MDIˆ = MECˆ (chứng minh trên )
DI = CE (chứng minh trên )DIMˆ = MCEˆ(chứng minh trên )=> ΔDIM = ΔECM (g-c-g)=> DM = ME ( cặp cạnh tương ứng )=> ĐPCM
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! ^_^