`b)` Xét `triangleABM` và `triangleACK` có:

`hat{A}` chung

`hat{AMB} = hat{AKC} = 90^@`

`=> triangleABM` $\backsim$ `triangleACK  ( g . g )`

`=> ( AM )/( AB) = ( AK)/( AC )  và   hat{A}` chung

`=> triangleAKM` $\backsim$ `triangleACB  ( c . g . c )` ( đpcm )

`c) CMTT câu `b => triangleCMD` $\backsim$ `triangleCBA  ( c . g . c )`

`=> hat{CMD} = hat{CBA}  ( 1 )`

`triangleAMK` $\backsim$ `triangleABC  ( c . g . c ) => hat{AMK} = hat{ABC}  ( 2 )`

`( 1 ) ; (2 ) => hat{AMK} = hat{DMC}  ( 3 )`

Lại có: `hat{AMK} + hat{KMB} = 90^@ ; hat{BMD} + hat{DMC} = 90^@  ( 4 )`

`( 3 ) ; ( 4 ) => hat{KMB} = hat{BMD} => MH` là phân giác `hat{KMD}` ( đpcm )