a) *Xét \Delta HADΔHAD và \Delta HCBΔHCB 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.⎩⎪⎪⎨⎪⎪⎧AH=HC(gt)AHD=CHB(đoˆˊi.đỉnh)BH=HD(gt)
\Rightarrow\Delta HAD=\Delta HCB\left(c-g-c\right)⇒ΔHAD=ΔHCB(c−g−c)
b) *Xét \Delta AHBΔAHB và \Delta CHDΔCHD 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.⎩⎪⎪⎨⎪⎪⎧AH=HC(gt)AHB=CHD(đoˆˊi.đỉnh)BH=HD(gt)
\Rightarrow\Delta AHB=\Delta CHD\left(c-g-c\right)⇒ΔAHB=ΔCHD(c−g−c)
\Rightarrow\widehat{HAB}=\widehat{HCD}⇒HAB=HCD (hai góc tương ứng)
Mà \widehat{HAB}HAB và \widehat{HCD}HCD ở vị trí so le trong
\Rightarrow AB//CD⇒AB//CD
c) *Xét \Delta AHMΔAHM và \Delta CHNΔCHNcó:
\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.⎩⎪⎪⎨⎪⎪⎧AH=HC(gt)AHM=CHN(đoˆˊi.đỉnh)HAM=HCN(cmt)
\Rightarrow\Delta AHM=\Delta CHN\left(g-c-g\right)⇒ΔAHM=ΔCHN(g−c−g)
\Rightarrow MH=HN⇒MH=HN (hai cạnh tương ứng)
*Xét \Delta CMHΔCMH và \Delta ANHΔANH 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.⎩⎪⎪⎨⎪⎪⎧CH=AH(gt)MHC=NHA(đoˆˊi.đỉnh)MH=HN(cmt)
\Rightarrow\Delta CMH=\Delta ANH\left(c-g-c\right)⇒ΔCMH=ΔANH(c−g−c)
