Câu `2 :`

Kẻ tia `Cx //// a => Cx //// b`

Ta có `: hat{A_2} + hat{ACx} = 180^0 ( 2` góc trong cùng phía `)`

`=> 115^0 + hat{ACx} = 180^0`

`=> hat{ACx} = 180^0 - 115^0`

`=> hat{ACx} = 65^0`

Lại có `: hat{B_1} = hat{BCx} = 25^0 ( 2` góc so le trong `)`

Có `: hat{ACB} = hat{ACx} + hat{BCx}`

`=> hat{ACB} = 65^0 + 25^0`

`=> hat{ACB} = 90^0`

`=> AC ⊥ CB (` Điều phải chứng minh `)`

Câu `3 :`

Kẻ tia `Cy //// a => Cy //// b`

Ta có `: hat{A_1} + hat{ACy} = 180^0 ( 2` góc trong cùng phía `)`

Lại có `: hat{B_1} + hat{BCy} = 180^0 ( 2` góc trong cùng phía `)`

Mà `hat{A_1} = hat{B_1} => hat{ACy} = hat{BCy} = 1/2 hat{ACB} = 1/2 x`

Vì `hat{A_1} + hat{ACy} = 180^0 => x + 1/2 x = 180^0`

`=> x ( 1 + 1/2 ) = 180^0`

`=> 3/2 x = 180^0`

`=> x = 180^0 : 3/2`

`=> x = 120^0`

Vậy `, x = 120^0 .`

`2)` Từ `C` kẻ $Cz // a$

`=>` $\widehat{A2}$ $+$ $\widehat{C1}$ $= 180^o ( 2$ góc $TCPBN )$

`=>` $\widehat{C1}$ $= 180^o - 115^o = 65^o$

Ta có $: a // b ( gt )$

Mà $Cz // a ($ cách dựng $)$

`=>` $Cz // b ( 3$ đường thẳng $// )$ 

`=>` $\widehat{B1}$ $=$ $\widehat{C2}$ $( 2$ góc sole trong $)$

`=>` $\widehat{C2}$ $= 25^o$

Ta có `:` $\widehat{C1}$ $+$ $\widehat{C2}$ $=$ $\widehat{ACB}$

`=>` $\widehat{ACB}$ `= 65^o + 25^o = 90^o`

`=> AC` $\bot$ $AB$

`3)` 

Từ `C` kẻ $Cz // a$

`=>` $\widehat{C1}$ $+$ $\widehat{A1}$ $= 180^o ( 2$ góc $TCPBN ) (1)$

Ta có $: a // b ( gt )$

Mà $Cz // a ($ cách dựng $)$

`=>` $Cz // b ( 3$ đường thẳng $// )$

`=>` $\widehat{C2}$ $+$ $\widehat{B1}$ $= 180^o ( 2$ góc $TCPBN ) (2)$

Mà $\widehat{A1}$ $=$ $\widehat{B1} ( gt ) (3)$

Từ $(1) ; (2) ; (3)$

`=>` $\widehat{C1}$ $=$ $\widehat{C2}$

`=>` $\widehat{C1}$ $=$ $\widehat{C2}$ `= 1/2ACB` 

`=>` $\widehat{A1}$ $+$ $\widehat{C1}$ $+$ $\widehat{B1}$ $+$ $\widehat{C2}$ $= 180^o + 180^o$

Hay `x + 1/2x + x + 1/2x = 360^o`

`=> 1x + 1/2x + 1x + 1/2x = 360^o`

`=> x( 1 + 1 + 1/2 + 1/2 ) = 360^o`

`=> 3x = 360^o`

`=> x = 360^o : 3`

`=> x = 120^o`