Xét ΔAEF và ΔDCF có

AE=DC

góc EAF=góc CDF

AF=DF

=>ΔAEF=ΔDCF

=>FE=CF

Xét ΔDCF và ΔBEC có

DC=BE

góc CDF=góc EBC

DF=BC

=>ΔDCF=ΔBEC

=>CF=CE

=>CF=CE=FE

=>ΔCEF đều

\(\widehat{ABE}=\widehat{ABC}+\widehat{CBE}=\widehat{ABC}+60^0\)    (do tam giác BCE đều)

\(\widehat{FDA}=\widehat{ADC}+\widehat{CDF}=\widehat{ADC}+60^0\) (do tam giác DFC đều)

 ABCD là hình bình hành  =>   \(\widehat{ABC}=\widehat{AD}C\)

suy ra:   \(\widehat{ABE}=\widehat{FDA}\)

Xét  \(\Delta ABE\)và     \(\Delta FDA\)có:

    \(AB=FD\)  (cùng bằng DC)

   \(\widehat{ABE}=\widehat{FDA}\) (cmt)

  \(BE=DA\)  (cùng bằng BC)

suy ra:   \(\Delta ABE=\Delta FDA\)  (c.g.c)

\(\Rightarrow\)\(AE=AF\)   (1)

Ta có:   \(\widehat{FCE}=360^0-\widehat{DCF}-\widehat{BCE}-\widehat{BCD}\)

                         \(=360^0-60^0-60^0-\widehat{BCD}\)

                         \(=240^0-\widehat{BCD}\)

                        \(=240^0-\left(180^0-\widehat{ABC}\right)=60^0+\widehat{ABC}\)

suy ra:   \(\widehat{FCE}=\widehat{ABE}\)

dễ dàng c/m:  \(\Delta ABE=\Delta FCE\)  (c.g.c)

\(\Rightarrow\)\(AE=FE\)       (2)

Từ (1) và (2) suy ra:   \(AF=FE=EA\)

hay    \(\Delta AEF\)đều

\(\Rightarrow\)\(\widehat{EAF}=60^0\)

mk lm dc rui nhug dù sao cux thanks bn

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Ta có:

∠ (BAD) + ∠ ∠ (ADC) = 180 0  (hai góc trong cùng phía bù nhau)

⇒  ∠ (ADC) =  180 0  -  ∠ (BAD) =  180 0  – α

∠ (CDF) =  ∠ (ADC) +  ∠ (ADF) =  180 0  - α 2 + 60 0 = 240 0 - α

Suy ra:  ∠ (CDF) =  ∠ (EAF)

Xét  ∆ AEF và ∆ DCF: AF = DF ( vì  ∆ ADF đều)

AE = DC (vì cùng bằng AB)

∠ (CDF) =  ∠ (EAF) (chứng minh trên)

Do đó:  ∆ AEF =  ∆ DCF (c.g.c) ⇒ EF = CF (1)

∠ (CBE) =  ∠ (ABC) + 60 0 = 180 0 - α + 60 0 = 240 0 - α

Xét ΔBCE và ΔDFC: BE = CD ( vì cùng bằng AB)

∠ (CBE) =  ∠ (CDF) = 240 0 - α

BC = DF (vì cùng bằng AD)

Do đó  ∆ BCE =  ∆ DFC (c.g.c) ⇒ CE = CF (2)

Từ (1) và (2) suy ra: EF = CF = CE

Vậy  ∆  ECF đều.

Bạn tự vẽ hình nhé!
Giải

a) Ta có:

\(\widehat{EAF}+\widehat{EAB}+\widehat{BAD}+\widehat{DAF}=360^0\)

\(\Rightarrow\widehat{EAF}+60^0+60^0+110^0=360^0\)

\(\Rightarrow\widehat{EAF}=130^o\)

b) Vì ABCD là hình bình hành nên:

\(\widehat{BAD}+\widehat{ADC}=180^o\)

\(110^o+\widehat{ADC}=180^o\)

\(\Rightarrow\widehat{ADC}=70^o\)

\(\Rightarrow\widehat{CDF}=\widehat{ADC}+\widehat{ADF}=70^o+60^o=130^o\)

Xét \(\Delta\)EAF và \(\Delta\)CDF có:\(\hept{\begin{cases}AE=DC\left(=AB\right)\\AF=DF\\\widehat{EAF}=\widehat{CDF}=130^o\end{cases}\Rightarrow\Delta EAF=\Delta CDF\left(cgc\right)}\)

c) Ta có: \(\Delta EAF=\Delta CDF\left(cmt\right)\Rightarrow EF=CF\)

Tương tự cũng có: \(\Delta CDF=\Delta EBC\left(cgc\right)\Rightarrow CF=EC\)

\(\Rightarrow\Delta\)EFC là tam giác đều (đpcm)

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Tính góc EAF 

EAF^ = 360* - (DAF^ + BAD^ + BAE^) = 360* - (60* + a + 60*) = 240* - a (1) 

b) Chứng minh rằng tam giác CEF là tam giác đều 

ABC^ = ADC^ = 180* - a 

=> CDF^ = ADC^ + ADF^ = 180* - a + 60* = 240* - a (2) 

CBE^ = ABC^ + ABE^ = 180* - a + 60* = 240* - a (3) 

AF = DF = AD = BC (4) 

CD = AB = BE = AE (5) 

(1) (2) (3) (4) và (5) => Δ CDF = ΔEBC = Δ EAF ( c.g.c) 

=> CF = CE = EF => CEF là tam giác đều

a, tính gócEAF

AEF^= 360- (DAF^+BAD^+BAE^)= 360 -(60+a+60)= 240-a (1)

b, chứng minh rằng tam giác CEF là tam giác đều

ABC^= ADC^=180 -a

=>CDF^=ADC^+ADF^=180-a+60=240-a (2)

CBE^=ABC^+ABE^=180-a+60=240-a (3)

AF=DF=AD=BC (4)

CD=AB=AE=BE (5)

Từ (1) (2) (3) (4) (5)=> tam giácCDF= tam giác EBC= tam giác EAF (c.g.c)

=>CF= CE= EF=> tam giác CEF ĐỀU