a: Xét ΔABH và ΔACH có
AB=AC
AH chung
BH=CH
Do đó: ΔABH=ΔACH
\(c,\left\{{}\begin{matrix}AI=IH\\\widehat{AIM}=\widehat{HIM}=90^0\\MI.chung\end{matrix}\right.\Rightarrow\Delta AMI=\Delta HMI\left(c.g.c\right)\\ \Rightarrow\widehat{MHA}=\widehat{MAH}\)
Mà \(\widehat{MAH}=\widehat{HAC}\left(\Delta ABH=\Delta ACH\right)\Rightarrow\widehat{MHA}=\widehat{HAC}\left(4\right)\)
\(\Delta ABH=\Delta ACH\Rightarrow\widehat{AHB}=\widehat{AHC}\)
Mà \(\widehat{AHB}+\widehat{AHC}=180^0\Rightarrow\widehat{AHB}=\widehat{AHC}=90^0\Rightarrow AH\perp BC\)
\(\left\{{}\begin{matrix}\widehat{HAC}+\widehat{ACH}=90^0\left(\Delta AHC\perp H\right)\\\widehat{ACH}+\widehat{HCO}=\widehat{ACO}=90^0\end{matrix}\right.\Rightarrow\widehat{HAC}=\widehat{HCO}\left(1\right)\\ \left\{{}\begin{matrix}\widehat{OHK}+\widehat{HOK}=90^0\left(\Delta HOK\perp K\right)\\\widehat{HOK}+\widehat{HCO}=90^0\left(\Delta HOC\perp H\right)\end{matrix}\right.\Rightarrow\widehat{OHK}=\widehat{HCO}\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\widehat{HAC}=\widehat{OHK}\left(3\right)\)
\(\left(3\right)\left(4\right)\Rightarrow\widehat{MHA}=\widehat{OHK}\)
Mà 2 góc này ở vị trí đối đỉnh và A,H,O thẳng hàng nên M,H,K thẳng hàng
