Xét \(\Delta ADM\) và \(\Delta AEM\) có :
\(\widehat{DAM}=\widehat{EAM}\) (gt)
\(\widehat{ADM}=\widehat{AEM}\) (\(=90^0\))
AM : cạnh chung
\(\Rightarrow\Delta ADM=\Delta AEM\) (\(ch-gn\))
Xét \(\Delta BDM\) và \(\Delta CEM\) có :
BM = MC (gt)
DM = EM (\(\Delta ADM=\Delta AEM\))
Vì \(\Delta BDM\perp D\)
\(\Rightarrow BD^2=MB^2-DM^2\)(định lí Pitago)
Vì \(\Delta CEM\perp E\)
\(\Rightarrow EC^2=MC^2-EM^2\)
Mà DM = ME , BM = MC
\(\Rightarrow MB^2-DM^2\) = \(MC^2-EM^2\)
\(\Rightarrow BD^2=EC^2\)
\(\Rightarrow BD=EC\)
\(\Rightarrow\Delta BDM=\Delta CEM\) (c . c . c)
Xét \(\Delta ABM\) và \(\Delta ACM\) có :
\(\widehat{BAM}=\widehat{CAM}\) (gt)
AM : cạnh chung
Ta có : BD = CE (\(\Delta BDM=\Delta CEM\))
Mà AD = AE (\(\Delta ADM=\Delta AEM\))
\(\Rightarrow AD+BD=AE+CE\)
\(\Rightarrow BA=AC\)
\(\Rightarrow\Delta ABM=\Delta ACM\) (c . g . c)