\(a,\left\{{}\begin{matrix}DE=DF\\AE=AF\\AD\text{ chung}\end{matrix}\right.\Rightarrow\Delta DEA=\Delta DFA\left(c.c.c\right)\\ b,\left\{{}\begin{matrix}DB=FB\\\widehat{DBK}=\widehat{ABF}\left(đđ\right)\\\widehat{DKB}=\widehat{BAF}\left(\text{so le trong}\right)\end{matrix}\right.\Rightarrow\Delta DBK=\Delta FBA\left(g.c.g\right)\\ \Rightarrow DK=AF\\ c,\Delta DEA=\Delta DFA\Rightarrow\widehat{DAE}=\widehat{DAF}\\ \Rightarrow AD\perp EF\\ \Rightarrow AD\perp DK\left(EF\text{//}DK\right)\\ \left\{{}\begin{matrix}\widehat{DAE}=\widehat{ADK}=90^0\\DK=AE\left(=AF\right)\\AD\text{ chung}\end{matrix}\right.\Rightarrow\Delta DAE=\Delta ADK\left(c.g.c\right)\\ \Rightarrow\widehat{DKB}=\widehat{E}\\ \Rightarrow\widehat{BAF}=\widehat{E}\\ \text{Mà 2 góc này ở vị trí đồng vị nên }AK\text{//}DE\)
\(d,\left\{{}\begin{matrix}\widehat{KEA}=\widehat{EKD}\\\widehat{MDK}=\widehat{MAE}=90^0\end{matrix}\right.\Rightarrow180^0-\widehat{KEA}-\widehat{MAE}=180^0-\widehat{EKD}-\widehat{MDK}\\ \Rightarrow\widehat{DMK}=\widehat{EMA}\\ \text{Mà 2 góc này ở vị trí đối đỉnh và }E,M,K\text{ thẳng hàng}\\ \text{Vậy }M,D,A\text{ thẳng hàng}\)
