`a)` Xét `\triangle HAB` và `\triangle KCA`

`{(AH = KC  \text{(gt)}),(AB = AC  \text{(gt)}),(HB = AK  \text{(gt)}):}`

`=> \triangle HAB = \triangle KCA  (c-c-c)`

`=> \hat{HBA} = \hat{KAC}`

`b)` Ta có: `\hat{HAB} + \hat{KAC} = 180^@ - \hat{BAC} = 180^@ - 90^@ = 90^@`

mà `\hat{HBA} = \hat{KAC}  (cmt)`

`=> \hat{HAB} + \hat{HBA} = 90^@`

`=> \hat{BHA} = 90^@`

`=> BH \bot AH`