a) Xét \(\Delta MAC,\Delta MDB\) có :
\(\left\{{}\begin{matrix}MA=MD\left(gt\right)\\\widehat{AMC}=\widehat{BMD}\left(\text{Đối đỉnh}\right)\\MC=MB\left(\text{AM là trung tuyến}\right)\end{matrix}\right.\)
=> \(\Delta MAC=\Delta MDB\left(c.g.c\right)\)
b) Xét \(\Delta BAC,\Delta DBA\) có :
\(\left\{{}\begin{matrix}BD=AC\left(\text{Suy ra từ câu a}\right)\\\widehat{BDA}=\widehat{ACB}\left(\text{Suy ra từ câu a}\right)\\AB:Chung\end{matrix}\right.\)
=> \(\Delta BAC=\Delta DBA\left(c.g.c\right)\)
=> \(\widehat{BAC}=\widehat{DBA}=90^o\) (2 góc tương ứng)
=> \(AB\perp BD\left(đpcm\right)\)
c) Từ \(\Delta BAC=\Delta DBA\left(c.g.c\right)\) suy ra :
\(BC=AD\) (2 cạnh tương ứng)
Mà : \(AM=\dfrac{AD}{2}\)
\(\Rightarrow AM=\dfrac{BC}{2}\)
=> đpcm.
