$\text{a) Vì BD = BA → ΔBAD cân tại B →}$ $\widehat{DAB}$ `=` $\widehat{D_1}$
$\text{Có:}$ $\widehat{H_2}$ `+` $\widehat{D_1}$ `+` $\widehat{A_2}$ `=` `180^o`
$\text{Hay:}$ `90^o` `+` $\widehat{D_1}$ `+` $\widehat{A_2}$ `=` `180^o`
`->` $\widehat{D_1}$ `+` $\widehat{A_2}$ `=` `180^o` `-` `90^o` `=` `90^o`
$\text{Lại có:}$ $\widehat{DAB}$ `+` $\widehat{A_3}$ `=` $\widehat{A}$ `=` `90^o`
$\text{Vì}$ $\widehat{D_1}$ `+` $\widehat{A_2}$ `=` $\widehat{DAB}$ `+` $\widehat{A_3}$ `=` `90^o`
$\text{Mà}$ $\widehat{DAB}$ `=` $\widehat{D_1}$
`->` $\widehat{A_2}$ `=` $\widehat{A_3}$
$\text{Xét ΔADH và ΔADE, ta có:}$
$\widehat{H_2}$ `=` $\widehat{E_1}$ `=` `90^o`
$\text{Chung AD}$
$\widehat{A_2}$ `=` $\widehat{A_3}$
`->` $\text{ΔADH = ΔADE (cạnh huyền - góc nhọn)}$
$\text{b) Vì}$ $\widehat{E_2}$ $\text{vuông tại E}$
`->` $\text{Cạnh DC lớn nhất}$
$\text{Có: BD + AE + DC > BD + AE + HC (Vì DC < HC)}$
$\text{Mà BD + DC = BC; AE + HC = AC}$
`->` $\text{AE + BC > BD + AC}$
$\text{Mà BD = BA}$
`->` $\text{AE + BC > AB + AC}$
$\text{Vì ΔADH = ΔADE (đcmt) → AH = AE}$
`->` $\text{AH + BC > AB + AC}$
