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24 tháng 2 2019

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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! ^_^

12 tháng 9

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

30 tháng 1 2021

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)

9 tháng 12 2021

cái nịt

11 tháng 2 2016

Bạn nào đó giải hộ mik điii

Mik đang bí bài này T^T

3 tháng 1 2021

cặc

23 tháng 8

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

23 tháng 3

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