a/ Có `hat{BaC}=hat{ACB}`
`=>hat{BAM}=hat{BCN}`
`=>∆BAM~∆BCN`
`=>(BM)/(BN)=(BA)/(BC)`
`=>(BM)/(AB)=(BN)/(BC)`
`=>∆BMN=∆BAC`
`=>hat{BMN}=hat{BAC}=hat{BCA}`
Mà 2 góc này đồng vị.
`=>MN//BC`
b/ `(MN)/(AC)=(BN)/(AB)`
`(BN)/(NA)=(BC)/(AC)`
`=>(BN)/(AB)=(BC)/(BC+AC)`
`=>(MN)/(AC)=(BC)/(BC+AC)`
`=>MN=(ab)/(a+b)`