Chứng minh MH=MK biết MH vuông góc AB, MK vuông góc AC

a/ \(\Delta ABC\) cân tại A

\(\Leftrightarrow\left\{{}\begin{matrix}AB=AC\\\widehat{B}=\widehat{C}\end{matrix}\right.\)

Xét \(\Delta ABM;\Delta ACM\) có :

\(\left\{{}\begin{matrix}AB=AC\\\widehat{B}=\widehat{C}\\AMchung\end{matrix}\right.\)

\(\Leftrightarrow\Delta ABM=\Delta ACM\left(c-g-c\right)\)

\(\Leftrightarrow\widehat{AMB}=\widehat{AMC}\)

Mà \(\widehat{AMB}+\widehat{AMC}=180^0\left(kềbuf\right)\)

\(\Leftrightarrow\widehat{AMB}=\widehat{AMC}=\dfrac{180^0}{2}=90^0\)

\(\Leftrightarrow AM\perp BC\left(đpcm\right)\)

b/ Ta có :

\(MB=MC=\dfrac{BC}{2}\) (M là trung điểm của BC)

\(\Leftrightarrow MB=MC=\dfrac{10}{2}=5cm\)

Xét \(\Delta AMB\) có \(\widehat{AMB}=90^0\)

\(\Leftrightarrow AB^2=AM^2+MB^2\) (định lí Py ta go)

\(\Leftrightarrow13^2=AM^2+5^2\)

\(\Leftrightarrow AM^2=13^2-5^2\)

\(\Leftrightarrow AM^2=144\)

\(\Leftrightarrow AM=12cm\)

c/ Xét \(\Delta BHM;\Delta CKM\) có :

\(\left\{{}\begin{matrix}\widehat{BHM}=\widehat{CKM}=90^0\\MB=MC\\\widehat{B}=\widehat{C}\end{matrix}\right.\)

\(\Leftrightarrow\Delta BHM=\Delta CKM\left(ch-gn\right)\)

\(\Leftrightarrow MH=MK\)