Qua `M` kẻ đoạn thẳng `EF` sao cho `EF////Aa` Có: `Aa////Bb` Mà `EF////Aa` `=>Aa////EF////Bb` Ta có: `aA////EF` `=>hat{A_1}=hat{AMF}(text{hai góc so le trong})` Mà `hat{A_1}=45^o` `=>hat{AMF}=45^o` Ta có: `Bb////EF` `=>hat{B_2}=hat{FMB}(text{hai góc so le trong})` Mà `hat{B_2}=30^o` `=>hat{FMB}=30^o` Mà `hat{AMB}=hat{AMF}+hat{FMB}` `=>hat{AMB}=45^o +30^o` `=>hat{AMB}=75^o` Vậy `hat{AMB}=75^o`
