Bài làm:

`\text{a)}` Xét `\triangle\text{ABC},` ta có`:`

`\hat\text{BAC}+\hat\text{ABC}+\hat\text{ACB}=180^\text{o}` `(`Tổng `3` góc trong một tam giác`)`

Hay `\hat\text{BAC}+70^\text{o}+70^\text{o}=180^\text{o}`

`=>``\hat\text{BAC}=180^\text{o}-(70^\text{o}+70^\text{o})=180^\text{o}-140^\text{o}=40^\text{o}`

`\text{b)}` Ta có`:`

`\hat\text{BAC}+\hat{x\text{AC}}=180^\text{o}` `(`Hai góc kề bù`)`

Hay `40^\text{o}+\hat{x\text{AC}}=180^\text{o}`

`=>``\hat{x\text{AC}}=180^\text{o}-40^\text{o}=140^\text{o}`

Vì `\text{A}z` là tia phân giác của `\hat{x\text{AC}}`

Nên `\hat{x\text{A}z}=\hat{z\text{AC}}=\hat{x\text{AC}}/2=140^\text{o}/2=70^\text{o}`

Suy ra`:` `\hat{z\text{AC}}=\hat\text{ACB}` `(=70^\text{o})`

Mà `\hat{z\text{AC}}=\hat\text{ACB}` ở bị trí so le trong

Nên `\text{A}z\text{//}\text{BC}` `(\text{đpcm})`

`\text{c)}`  Vì `\text{A}m` là tia phân giác của `\hat{z\text{AC}}`

Nên `\hat{m\text{A}z}=\hat{m\text{AC}}=\hat{z\text{AC}}/2=70^\text{o}/2=35^\text{o}`

`=>``\hat\text{EAC}=35^\text{o}`

Lại có`:`

`\hat\text{ACB}+\hat{\text{AC}y}=180^\text{o}` `(`Hai góc kề bù`)`

Hay `70^\text{o}+\hat{\text{AC}y}=180^\text{o}`

`=>``\hat{\text{AC}y}=180^\text{o}-70^\text{o}=110^\text{o}`

Vì `\text{C}t` là tia phân giác của `\hat{\text{AC}y}`

Nên `\hat{t\text{CA}}=\hat{t\text{C}y}=\hat{\text{AC}y}/2=110^\text{o}/2=55^\text{o}`

`=>``\hat\text{ACE}=55^\text{o}`

Xét `\triangle\text{AEC},` ta có`:`

`\hat\text{EAC}+\hat\text{ACE}+\hat\text{AEC}=180^\text{o}` `(`Tổng `3` góc trong một tam giác`)`

Hay `35^\text{o}+55^\text{o}+\hat\text{AEC}=180^\text{o}`

`=>``\hat\text{AEC}=180^\text{o}-(35^\text{o}+55^\text{o})=180^\text{o}-90^\text{o}=90^\text{o}`

Suy ra`:` `\triangle\text{AEC}` vuông tại `\text{E}`

`=>``\text{CE}``\bot``\text{AE}` `(\text{đpcm})`