Ta có: \(\Delta ABC\) vuông cân tại A
\(\Rightarrow\hept{\begin{cases}\widehat{BAC}=90^0\\AB=AC\\\widehat{ABC}=\widehat{ACB}=45^0\end{cases}}\)
Lại có: \(\hept{\begin{cases}\widehat{BAH}+\widehat{HAC}=90^0\\\widehat{KCA}+\widehat{HAC}=90^0\end{cases}}\)
\(\Rightarrow\widehat{BAH}=\widehat{KCA}\)
Xét \(\Delta ABH\) và \(\Delta CAK:\)
\(\hept{\begin{cases}\widehat{AHB}=\widehat{CKA}=90^0\\AB=AC\left(cmt\right)\\\widehat{BAH}=\widehat{KCA}\left(cmt\right)\end{cases}}\)
\(\Rightarrow\Delta ABH=\Delta CAK\left(ch+gn\right)\)
\(\Rightarrow AH=CK\)
Có: \(\hept{\begin{cases}AM⊥MB\\\widehat{ABM}=45^0\end{cases}}\)
\(\Rightarrow\widehat{MAB}=45^0=\widehat{ACM}\)
\(\Rightarrow\widehat{BAH}-\widehat{BAM}=\widehat{KCA}-\widehat{ACM}\)
\(\Rightarrow\widehat{HAM}=\widehat{KCM}\)
Ta lại có: \(\hept{\begin{cases}AM⊥MC\\\widehat{AMC}=45^0\end{cases}}\)
\(\Rightarrow\widehat{MAC}=45^0\)
\(\Rightarrow\Delta AMC\) vuông cân.\(\Rightarrow MA=MC\)
Xét \(\Delta AMH\) và \(\Delta CMK:\)
\(\hept{\begin{cases}AH=KC\left(cmt\right)\\\widehat{HAM}=\widehat{KCM}\left(cmt\right)\\AM=CM\left(cmt\right)\end{cases}}\)
\(\Rightarrow\Delta AMH=\Delta CMK\left(c.g.c\right)\)
\(\Rightarrow MK=MH.\)
