$\text{a) Vì}$ $\Delta{ABC}$ $\text{cân tại A}$
`->` $\text{AB = AC và}$ $\widehat{B}$ `=` $\widehat{C}$
$\text{Xét}$ $\Delta{ABM}$ $\text{và}$ $\Delta{ACM}$$\text{, ta có:}$
$\text{AB = AC}$
$\text{MB = MC (Vì M là trung điểm của BC)}$
$\text{Chung AM}$
`->` $\Delta{ABM}$ $\text{=}$ $\Delta{ACM}$ $\text{(c.c.c)}$
$\text{b) Xét}$ $\Delta{MHB}$ $\text{và}$ $\Delta{MKC}$$\text{, ta có:}$
$\widehat{B}$ `=` $\widehat{C}$
$\text{MB = MC}$
$\widehat{MKC}$ `=` $\widehat{MHB}$ `= 90^o`
`->` $\Delta{MHB}$ $\text{=}$ $\Delta{MKC}$ $\text{(cạnh huyền - góc nhọn)}$
`->` $\text{BH = CK (2 cạnh tương ứng)}$
$\text{c) Vì}$ $\Delta{MHB}$ $\text{=}$ $\Delta{MKC}$ $\text{(đcmt)}$
`->` $\widehat{CMK}$ `=` $\widehat{BMH}$ $\text{(2 góc tương ứng) (1)}$
$\text{Vì}$ $\left \{ {{MK ⊥ AC} \atop {BP ⊥ AC}} \right.$ `->` $\text{MK // BP}$
$\text{Vì MK // BP}$
`->` $\widehat{IBM}$ `=` $\widehat{CMK}$ $\text{(2 góc đồng vị) (2)}$
$\text{Từ (1) và (2) suy ra}$ $\widehat{CMK}$ `=` $\widehat{IBM}$`=` $\widehat{BMH}$
$\text{Vì}$ $\widehat{IBM}$`=` $\widehat{BMH}$
`->` $\Delta{IMB}$ $\text{cân tại I}$
$\boxed{\text{Selina}}$
