Bài 8: Đường tròn nội tiếp. Đường tròn ngoại tiếp

Lời giải:
\(S_{ABC}=\frac{AD.BC}{2}; S_{ABMC}=\frac{AM.BC}{2}\)

\(\Rightarrow \frac{S_{ABMC}}{S_{ABC}}=\frac{AM}{AD}\)

Hoàn toàn TT: \(\frac{S_{ABCN}}{S_{ABC}}=\frac{BN}{BE}; \frac{S_{ACBK}}{S_{ABC}}=\frac{CK}{CF}\)

Do đó:
\(\frac{AM}{AD}+\frac{BN}{BE}+\frac{CK}{CF}=\frac{S_{ABMC}+S_{ABCN}+S_{ACBK}}{S_{ABC}}\)

\(=\frac{S_{ABC}+S_{BMC}+S_{ABC}+S_{ANC}+S_{ABC}+S_{AKB}}{S_{ABC}}=3+\frac{S_{BMC}+S_{ANC}+S_{AKB}}{S_{ABC}}(*)\)

Lại có:

\(\widehat{MBD}=\widehat{MBC}=\widehat{MAC}\) (góc nội tiếp cùng chắn cung MC)

\(=\widehat{HAE}=90^0-\widehat{AHE}=90^0-\widehat{BHD}=\widehat{HBD}\)

Xét tam giác $HBD$ và $MBD$ có:

\(\left\{\begin{matrix} \widehat{MBD}=\widehat{HBD}\\ \widehat{BDH}=\widehat{BDM}=90^0\end{matrix}\right.\) \(\Rightarrow \triangle HBD\sim \triangle MBD\)

\(\Rightarrow \frac{HD}{BD}=\frac{MD}{BD}\Rightarrow HD=MD\)

\(\Rightarrow S_{BHC}=\frac{HD.BC}{2}=\frac{MD.BC}{2}=S_{BMC}\)

Hoàn toàn TT: \(S_{AHC}=S_{ANC}; S_{AHB}=S_{AKB}\)

\(\Rightarrow S_{BMC}+S_{ANC}+S_{AKB}=S_{BHC}+S_{AHC}+S_{AHB}=S_{ABC}(**)\)

Từ \((1);(2)\Rightarrow \frac{AM}{AD}+\frac{BN}{BE}+\frac{CK}{CF}=3+\frac{S_{ABC}}{S_{ABC}}=4\) (đpcm)