`a,` Các cặp tam giác đồng dạng là :

\(\Delta AMN\sim\Delta ABC\) `(` vì \(MN\text{/}\text{/}BC\)  `)`

\(\Delta ABC\sim\Delta MBL\) `(` vì \(ML\text{/}\text{/}AC\)  `)`

\(\Delta AMN\sim\Delta MBL\)

`b,` * \(\Delta AMN\sim\Delta ABC\)  thì

\(\left\{{}\begin{matrix}\widehat{MAN}=\widehat{BAC}\\\widehat{AMN}=\widehat{ABC}\\\widehat{ANM}=\widehat{ACB}\end{matrix}\right.\)

\(\dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{MN}{BC}\)

* \(\Delta ABC\sim\Delta MBL\)  thì

\(\left\{{}\begin{matrix}\widehat{BAC}=\widehat{BML}\\\widehat{ABC}=\widehat{MBL}\\\widehat{ACB}=\widehat{MLB}\end{matrix}\right.\)

\(\dfrac{AB}{MB}=\dfrac{BC}{BL}=\dfrac{AC}{ML}\)

* \(\Delta AMN\sim\Delta MBL\)  thì

\(\left\{{}\begin{matrix}\widehat{MAN}=\widehat{BML}\\\widehat{AMN}=\widehat{MBL}\\\widehat{ANM}=\widehat{MLB}\end{matrix}\right.\)

\(\dfrac{AM}{MB}=\dfrac{AN}{ML}=\dfrac{MN}{BL}\)

a) ΔABC có MN // BC (M ∈ AB; N ∈ AC) ⇒ ΔAMN  ΔABC.

ΔABC có ML // AC (M ∈ AB; L ∈ BC) ⇒ ΔMBL  ΔABC

ΔAMN  ΔABC; ΔMBL  ΔABC ⇒ ΔAMN  ΔMBL.

b) ΔAMN  ΔABC có:

ΔMBL  ΔABC có:

ΔAMN  ΔMBL có:

a, Tam giác AMN ~ Tam giác ABC 

Tam giác MBL ~ tam giác ABC 

Tam giác AMN ~ tam giác MBL 

Kiếm trên mạng ấy, nhiều lắm đó bạn.

https://www.google.com/search?q=T%E1%BB%AB+%C4%91i%E1%BB%83m+M+thu%E1%BB%99c+c%E1%BA%A1nh+AB+c%E1%BB%A7a+tam+gi%C3%A1c+ABC+v%E1%BB%9Bi+AM+%3D+1+2+12+MB%2C+k%E1%BA%BB+c%C3%A1c+tia+song+song+v%E1%BB%9Bi+AC+v%C3%A0+BC%2C+ch%C3%BAng+c%E1%BA%AFt+BC+v%C3%A0+AC+l%E1%BA%A7n+l%C6%B0%E1%BB%A3t+t%E1%BA%A1i+L+v%C3%A0+N+a%2C+N%C3%AAu+t%E1%BA%A5t+c%E1%BA%A3+c%C3%A1c+c%E1%BA%B7p+tam+gi%C3%A1c+%C4%91%E1%BB%93ng+d%E1%BA%A1ng+b%2C+%C4%91%E1%BB%91i+v%E1%BB%9Bi+m%E1%BB%97i+c%E1%BA%B7p+tam+gi%C3%A1c+%C4%91%E1%BB%93ng+d%E1%BA%A1ng%2C+h%C3%A3y+vi%E1%BA%BFt+c%C3%A1c+c%E1%BA%B7p+g%C3%B3c+b%E1%BA%B1ng+nhau+v%C3%A0+t%E1%BB%89+s%E1%BB%91+%C4%91%E1%BB%93ng+d%E1%BA%A1ng+t%C6%B0%C6%A1ng+%E1%BB%A9ng&oq=T%E1%BB%AB+%C4%91i%E1%BB%83m+M+thu%E1%BB%99c+c%E1%BA%A1nh+AB+c%E1%BB%A7a+tam+gi%C3%A1c+ABC+v%E1%BB%9Bi+AM+%3D%C2%A0++1+2+12+MB%2C+k%E1%BA%BB+c%C3%A1c+tia+song+song+v%E1%BB%9Bi+AC+v%C3%A0+BC%2C+ch%C3%BAng+c%E1%BA%AFt+BC+v%C3%A0+AC%C2%A0l%E1%BA%A7n+l%C6%B0%E1%BB%A3t+t%E1%BA%A1i+L+v%C3%A0+N++a%2C+N%C3%AAu+t%E1%BA%A5t+c%E1%BA%A3+c%C3%A1c+c%E1%BA%B7p+tam+gi%C3%A1c+%C4%91%E1%BB%93ng+d%E1%BA%A1ng%C2%A0++b%2C+%C4%91%E1%BB%91i+v%E1%BB%9Bi+m%E1%BB%97i%C2%A0c%E1%BA%B7p+tam+gi%C3%A1c+%C4%91%E1%BB%93ng+d%E1%BA%A1ng%2C+h%C3%A3y+vi%E1%BA%BFt+c%C3%A1c+c%E1%BA%B7p+g%C3%B3c+b%E1%BA%B1ng+nhau+v%C3%A0+t%E1%BB%89+s%E1%BB%91+%C4%91%E1%BB%93ng+d%E1%BA%A1ng+t%C6%B0%C6%A1ng+%E1%BB%A9ng&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8

a) MN // BC => ∆AMN ∽ ∆ABC

ML // AC => ∆MBL ∽ ∆ABC

và ∆AMN ∽ ∆MLB

b)

∆AMN ∽ ∆ABC có:

$\widehat{AMN}$ = $\widehat{ABC}$; $\widehat{ANM}$ = $\widehat{ACB}$

$\frac{AM}{AB}$= $\frac{1}{3}$

∆MBL ∽ ∆ABC có:

$\widehat{MBL}$ = $\widehat{BAC}$, $\hat{B}$ chung, $\widehat{MLB}$ = $\widehat{ACB}$

$\frac{MB}{AB}$= $\frac{2}{3}$

∆AMN ∽ ∆MLB có:

$\widehat{MAN}$ = $\widehat{BML}$, $\widehat{AMN}$ = $\widehat{MBL}$, $\widehat{ANM}$ =

a) MN // BC => ∆AMN ∽ ∆ABC

ML // AC => ∆MBL ∽ ∆ABC và ∆AMN ∽ ∆MLB

b) ∆AMN ∽ ∆ABC có:

ˆAMNAMN^ = ˆABCABC^; ˆANMANM^ = ˆACBACB^

$\frac{AM}{AB}$= $\frac{1}{3}$

 ∆MBL ∽ ∆ABC có: 

ˆMBLMBL^ = ˆBACBAC^, ˆBB^ chung, ˆMLBMLB^ = ˆACBACB^

$\frac{MB}{AB}$= $\frac{2}{3}$

∆AMN ∽ ∆MLB có:

ˆMANMAN^ = ˆBMLBML^, 

a) ΔAMN∼ΔABC

ΔBML∼ΔBAC

b) Ta có: ΔAMN∼ΔABC(cmt)

nên \(\widehat{AMN}=\widehat{ABC}\); \(\widehat{ANM}=\widehat{ACB}\); \(\widehat{A}\) chung và \(\dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{MN}{BC}\)

Ta có: ΔBML∼ΔBAC(cmt)

nên \(\widehat{BML}=\widehat{BAC}\); \(\widehat{BLM}=\widehat{BCA}\); \(\widehat{B}\) chung và \(\dfrac{BM}{BA}=\dfrac{ML}{AC}=\dfrac{BL}{BC}\)

Bài này là: Bài 27 trang 72 Toán 8 Tập 2 đúng không bạn 

a) \(\Delta ABC\) có \(MN\) // \(BC\) \(\left(M\in AB;N\in AC\right)\Rightarrow\Delta AMN\sim\Delta ABC\) (định lí)

\(\Delta ABC\) có \(ML\) // \(AC\) \(\left(M\in AB;L\in BC\right)\Rightarrow\Delta MBL\sim\Delta ABC\) (định lí)

Vì \(\Delta AMN\sim\Delta ABC\) và \(\Delta MBL\sim\Delta ABC\)

\(\Rightarrow\Delta AMN\sim\Delta MBL\)

b) Xét \(\Delta AMN\sim\Delta ABC\) có:

\(\widehat{A}\) chung

\(\widehat{AMN}=\widehat{B};\widehat{ANM}=\widehat{C}\)

\(\dfrac{AM}{AB}=\dfrac{AN}{AC}=\dfrac{MN}{BC}\)

Tỉ số đồng dạng : \(k=\dfrac{AM}{AB}=\dfrac{1}{2}\left(AM=\dfrac{1}{2}MB\right)\)

Xét \(\Delta MBL\sim\Delta ABC\) có:

\(\widehat{B}\) chung

\(\widehat{BML}=\widehat{A};\widehat{MLK}=\widehat{C}\)

\(\dfrac{BM}{BA}=\dfrac{BL}{BC}=\dfrac{ML}{AC}\)

Tỉ số đồng dạng: \(k'=\dfrac{BM}{BA}=\dfrac{2}{3}\)

Xét \(\Delta AMN\sim\Delta MBL\) có:

\(\widehat{AMN}=\widehat{B};\widehat{ANM}=\widehat{BLM};\widehat{A}=\widehat{BML}\)

\(\dfrac{AM}{MB}=\dfrac{AN}{ML}=\dfrac{MN}{BL}\)

Tỉ số đồng dạng: \(k''=\dfrac{AM}{MB}=\dfrac{1}{2}\)

a: Xét ΔAMI và ΔABC có

góc AMI=góc ABC

góc A chung

=>ΔAMI đồng dạng với ΔABC

Xét ΔBMN và ΔBAC có

góc B chung

góc BMN=góc BAC

=>ΔBMN đồng dạng với ΔBAC

=>ΔMBN đồng dạng với ΔABC

=>ΔMBN đồng dạng với ΔAMI

b: ΔAMI đồng dạng với ΔABC

=>AM/AB=AI/AC=MI/BC và góc AMI=góc ABC; góc AIM=góc ACB

ΔMBN đồng dạng với ΔABC

=>MB/BA=BN/BC=MN/AC và góc BMN=góc BAC; góc BNM=góc BCA

ΔAMI đồng dạng với ΔMBN

=>AM/MB=MI/BN=AI/MN và góc MAI=góc MBN; góc AMI=góc MBN; góc AIM=góc MNB