\(1,\left\{{}\begin{matrix}\widehat{AHD}=\widehat{BKC}\left(=90^0\right)\\AD=BC\left(hthang.cân.ABCD\right)\\\widehat{ADH}=\widehat{BCK}\left(hthang.cân.ABCD\right)\end{matrix}\right.\Rightarrow\Delta ADH=\Delta BCK\left(ch-gn\right)\\ 2,\Delta ADH=\Delta BCK\Rightarrow\widehat{ODC}=\widehat{OCD}\)
Mà \(\widehat{ODC}=\widehat{OBA}\left(so.le.trong\right);\widehat{OCD}=\widehat{OAB}\left(so.le.trong\right)\)
\(\Rightarrow\widehat{OBA}=\widehat{OAB}\Rightarrow\Delta OAB.cân.tại.O\Rightarrow OA=OB\\ \Rightarrow O\inđường.trung.trực.của.AB\left(1\right)\)
\(\widehat{IDC}=\widehat{ICD}\Rightarrow\Delta ICD.cân.tại.I\Rightarrow IC=ID\\ \Rightarrow ID-AD=IC-BC\Rightarrow IA=IB\\ \Rightarrow I\inđường.trung.trực.của.AB\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow OI.là.trung.trực.của.AB\)
\(c,\) Dễ thấy ABKH là hcn do \(AH//BK\left(\perp CD\right);AH=BK;\widehat{AHK}=90^0\)
\(\Rightarrow AB=HK\)
Ta có \(BK=\dfrac{AB+CD}{2}\Rightarrow2BK=AB+CD\)
\(\Rightarrow2BK=HK+\left(DH+HK+KC\right)=2HK+\left(DH+HC\right)\)
Mà \(DH=HC\left(\Delta AHD=\Delta BKC\right)\)
\(\Rightarrow2BK=2HK+2DH\Rightarrow BK=HK+DH=DK\)
Do đó \(\Delta BKD\) vuông cân
\(\Rightarrow\widehat{KBD}=\widehat{KDB}=45^0\)
Mà \(\widehat{KDB}.hay.\widehat{ODC}=\widehat{OCD}=\widehat{OAB}=\widehat{OBA}=45^0\left(câu.2\right)\)
