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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
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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! ^_^
a: Qua D, kẻ DK//AC(K∈BC)
DK//AC
=>\(\hat{DKB}=\hat{ACB}\) (hai góc đồng vị)
=>\(\hat{DBK}=\hat{DKB}\)
=>DB=DK
mà DB=CE
nên DK=CE
Xét ΔIKD và ΔICE có
\(\hat{IKD}=\hat{ICE}\) (hai góc so le trong, KD//CE)
KD=CE
\(\hat{IDK}=\hat{IEC}\) (hai góc so le trong, DK//EC)
Do đó: ΔIKD=ΔICE
=>ID=IE và IK=IC
ID=IE nên I là trung điểm của DE
Ta có: \(\hat{ABC}+\hat{ABF}=180^0\) (hai góc kề bù)
\(\hat{ACB}+\hat{BCE}=180^0\) (Hai góc kề bù)
mà \(\hat{ABC}=\hat{ACB}\)
nên \(\hat{ABF}=\hat{BCE}\)
Xét ΔFBD và ΔICE có
FB=IC
\(\hat{FBD}=\hat{ICE}\)
BD=CE
Do đó: ΔFBD=ΔICE
=>FD=IE
mà IE=ID
nên FD=ID
=>ΔFDI cân tại D
b: Xét ΔABC có \(\frac{AD}{AB}=\frac{AM}{AC}\)
nên DM//BC
c: Ta có: AB=AC
=>A nằm trên đường trung trực của BC(1)
Ta có: NB=NC
=>N nằm trên đường trung trực của BC(2)
Từ (1),(2) suy ra AN là đường trung trực của BC
a: Qua D, kẻ DK//AC(K∈BC)
DK//AC
=>\(\hat{DKB}=\hat{ACB}\) (hai góc đồng vị)
=>\(\hat{DKB}=\hat{DBK}\)
=>DK=DB
mà DB=CE
nên DK=CE
Xét ΔIKD và ΔICE có
\(\hat{IKD}=\hat{ICE}\) (hai góc so le trong, DK//CE)
DK=CE
\(\hat{IDK}=\hat{IEC}\) (hai góc so le trong, DK//CE)
Do đó: ΔIKD=ΔICE
=>ID=IE và IK=IC
ID=IE
=>I là trung điểm của DE
Ta có: \(\hat{ABC}+\hat{ABF}=180^0\) (hai góc kề bù)
\(\hat{ACB}+\hat{ICE}=180^0\) (hai góc kề bù)
mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)
nên \(\hat{ABF}=\hat{ICE}\)
Xét ΔDBF và ΔECI có
DB=EC
\(\hat{DBF}=\hat{ECI}\)
BF=EI
Do đó: ΔDBF=ΔECI
=>DF=EI
mà EI=DI
nên DF=DI
=>ΔDFI cân tại D
b: Xét ΔABC có \(\frac{AD}{AB}=\frac{AM}{AC}\)
nên DM//BC
c: Ta có: AB=AC
=>A nằm trên đường trung trực của BC(1)
Ta có: NB=NC
=>N nằm trên đường trung trực của BC(2)
Từ (1),(2) suy ra AN là đường trung trực của BC
a: Qua D, kẻ DK//AC(K∈BC)
DK//AC
=>\(\hat{DKB}=\hat{ACB}\) (hai góc đồng vị)
=>\(\hat{DKB}=\hat{DBK}\)
=>DK=DB
mà DB=CE
nên DK=CE
Xét ΔIKD và ΔICE có
\(\hat{IKD}=\hat{ICE}\) (hai góc so le trong, DK//CE)
DK=CE
\(\hat{IDK}=\hat{IEC}\) (hai góc so le trong, DK//CE)
Do đó: ΔIKD=ΔICE
=>ID=IE và IK=IC
ID=IE
=>I là trung điểm của DE
Ta có: \(\hat{ABC}+\hat{ABF}=180^0\) (hai góc kề bù)
\(\hat{ACB}+\hat{ICE}=180^0\) (hai góc kề bù)
mà \(\hat{ABC}=\hat{ACB}\) (ΔABC cân tại A)
nên \(\hat{ABF}=\hat{ICE}\)
Xét ΔDBF và ΔECI có
DB=EC
\(\hat{DBF}=\hat{ECI}\)
BF=EI
Do đó: ΔDBF=ΔECI
=>DF=EI
mà EI=DI
nên DF=DI
=>ΔDFI cân tại D
b: Xét ΔABC có \(\frac{AD}{AB}=\frac{AM}{AC}\)
nên DM//BC
c: Ta có: AB=AC
=>A nằm trên đường trung trực của BC(1)
Ta có: NB=NC
=>N nằm trên đường trung trực của BC(2)
Từ (1),(2) suy ra AN là đường trung trực của BC
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)