a) Tứ giác ABHE nội tiếp ( vì $\widehat{AEB}$ = $\widehat{AHB}$ = `90^o`
`=>` $\widehat{BEH}$ = $\widehat{BAH}$ = `1/2` $\widehat{BH}$
Mà $\widehat{ACB}$=$\widehat{BAH}$ ( = `90^o` - $\widehat{ABC}$)
`=>` $\widehat{BEH}$ = $\widehat{FCB}$
Xét $\triangle$BHE và $\triangle$BFC có:
B chung
$\widehat{BEH}$=$\widehat{BCF}$
`=>` $\triangle$DEH $\backsim$ $\triangle$BCF (gg)
`=>` $\dfrac{BE}{BC}$=$\dfrac{BH}{BF}$`=>` BE.BF = BF.BC
b) `AB^2` = `Ah^2` + `BH^2`
`AC^2` = `AH^2` + `CH^2`
`=>` $\dfrac{AB^{2}}{AC^{2}}$= $\dfrac{AH^{2} + BH^{2}}{AH^{2} + CH^{2}}$=$\dfrac{BH.CH+CH^{2}}{BH.CH+CH^{2}}$
`=>` $\dfrac{BH(BH+CH)}{CH(BH+CH)}$=$\dfrac{BH}{CH}$
