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)

a: Ta có: \(\hat{ABC}+\hat{DBK}=180^0\) (hai góc kề bù)

\(\hat{ACB}+\hat{ICE}=180^0\) (hai góc kề bù)

mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)

nên \(\hat{DBK}=\hat{ECI}\)

Xét ΔDBK và ΔECI có

DB=EC

\(\hat{DBK}=\hat{ECI}\)

BK=CI

Do đó: ΔDBK=ΔECI

b: ΔDBK=ΔECI

=>KD=EI

Qua D, kẻ DH//AC(H∈BC)

DH//AC

=>\(\hat{DHB}=\hat{ACB}\) (hai góc đồng vị)

mà \(\hat{DBH}=\hat{ACB}\) (ΔABC cân tại A)

nên \(\hat{DBH}=\hat{DHB}\)

=>DH=DB

mà DB=CE
nên DH=CE

Xét ΔIHD và ΔICE có

\(\hat{IDH}=\hat{IEC}\) (hai góc so le trong, DH//CE)

DH=CE
\(\hat{IHD}=\hat{ICE}\) (hai góc so le trong, DH//CE)

Do đó:ΔIHD=ΔICE
=>DH=CE

ΔIHD=ΔICE

=>DI=IE

mà IE=DK

nên DI=DK

=>ΔDIK cân tại D

c: ΔABC cân tại A

mà AO là đường cao

nên AO là phân giác của góc BAC

Xét ΔABO và ΔACO có

AB=AC

\(\hat{BAO}=\hat{CAO}\)

AO chung

Do đó: ΔABO=ΔACO

=>OB=OC

ΔABO=ΔACO

=>\(\hat{ABO}=\hat{ACO}\)

=>\(\hat{ACO}=90^0\)

=>OC⊥AE tại C

Xét ΔDBO vuông tại B và ΔECO vuông tại C có

DB=EC

OB=OC

Do đó: ΔDBO=ΔECO

a: Sửa đề: EK⊥BC

Ta có: \(\hat{DBH}=\hat{ACB}\) (ΔABC cân tại A)

\(\hat{ECK}=\hat{ACB}\) (hai góc đối đỉnh)

Do đó: \(\hat{DBH}=\hat{ECK}\)

Xét ΔDHB vuông tại H và ΔEKC vuông tại K có

DB=CE

\(\hat{DBH}=\hat{ECK}\)

Do đó: ΔDHB=ΔEKC

=>BH=CK

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

ABCDEMI

Từ D kẻ DI // AE

Vì ΔΔABC cân tại A nên BˆB^ = ACBˆACB^ (1)

Vì DI // AE => ACBˆACB^ = DIBˆDIB^ (đồng vị ) (2)

Từ (1) và (2) => BˆB^ = DIBˆDIB^

Trong ΔΔDIB có : BˆB^ = DIBˆDIB^ => ΔΔDIB cân tại D

=> DB = DI mà DB = CE (gt)

=> DI = CE

Vì DI // AE => MDIˆMDI^ =