a) *Xét \(\Delta HAD\) và \(\Delta HCB\) có:
\(\left\{{}\begin{matrix}AH=HC\left(gt\right)\\\widehat{AH\text{D}}=\widehat{CHB}\left(\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\BH=HD\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta HAD=\Delta HCB\left(c-g-c\right)\)
b) *Xét \(\Delta AHB\) và \(\Delta CHD\) có:
\(\left\{{}\begin{matrix}AH=HC\left(gt\right)\\\widehat{AHB}=\widehat{CHD}\left(\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\BH=HD\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta AHB=\Delta CHD\left(c-g-c\right)\)
\(\Rightarrow\widehat{HAB}=\widehat{HCD}\) (hai góc tương ứng)
Mà \(\widehat{HAB}\) và \(\widehat{HCD}\) ở vị trí so le trong
\(\Rightarrow AB//CD\)
c) *Xét \(\Delta AHM\) và \(\Delta CHN\)có:
\(\left\{{}\begin{matrix}AH=HC\left(gt\right)\\\widehat{AHM}=\widehat{CHN}\left(\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\\widehat{HAM}=\widehat{HCN}\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta AHM=\Delta CHN\left(g-c-g\right)\)
\(\Rightarrow MH=HN\) (hai cạnh tương ứng)
*Xét \(\Delta CMH\) và \(\Delta ANH\) có:
\(\left\{{}\begin{matrix}CH=AH\left(gt\right)\\\widehat{MHC}=\widehat{NHA}\left(\text{đ}\text{ối}.\text{đ}\text{ỉnh}\right)\\MH=HN\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta CMH=\Delta ANH\left(c-g-c\right)\)
