Question

mng giúp e zới ạ!!!

Answer 1

a) Ta có:

\(\widehat{BDM}+\widehat{DBM}+\widehat{BMD}=180^0\)(Tổng 3 góc trong tam giác)

\(\widehat{BMD}+\widehat{DME}+\widehat{EMC}=180^0\)

Mà \(\widehat{DBM}=\widehat{DME}=60^0\)

\(\Rightarrow\widehat{BDM}=\widehat{EMC}\)

Xét tam giác BDM và tam giác CME có:

\(\left\{{}\begin{matrix}\widehat{BDM}=\widehat{EMC}\left(cmt\right)\\\widehat{DBM}=\widehat{ECM}=60^0\end{matrix}\right.\)

\(\Rightarrow\Delta BDM\sim\Delta CME\left(g.g\right)\Rightarrow\dfrac{BD}{BM}=\dfrac{CM}{CE}\Rightarrow BD.CE=BM.CM\)

Mà \(BM=CM=\dfrac{BC}{2}\)(M là trung điểm BC) 

\(\Rightarrow BD.CE=\left(\dfrac{BC}{2}\right)^2=\dfrac{BC^2}{4}=\dfrac{a^2}{4}\left(đpcm\right)\)

b) Ta có: \(\Delta BDM\sim\Delta CME\left(cmt\right)\Rightarrow\dfrac{BD}{MD}=\dfrac{CM}{ME}=\dfrac{BM}{ME}\)

Xét tam giác MBD và tam giác EMD có:

\(\dfrac{BD}{MD}=\dfrac{BM}{ME}\left(cmt\right)\)

\(\widehat{DBM}=\widehat{DME}=60^0\)

\(\Rightarrow\Delta MBD\sim\Delta EMD\left(c.g.c\right)\)

Ta có: \(\left\{{}\begin{matrix}\Delta MBD\sim\Delta ECM\left(cmt\right)\\\Delta MBD\sim\Delta EMD\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta ECM\sim\Delta EMD\left(đpcm\right)\)