a. Xét \(\Delta ABC\) có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (Tổng 3 góc của tam giác)
\(\Rightarrow\widehat{A}=180^o-\widehat{B}-\widehat{C}=180^o-60^o-40^o=80^o\)
b. Xét \(\Delta BED\) và \(\Delta BEC\) có:
\(BD=BC\left(g.t\right)\)
\(\widehat{DBE}=\widehat{CBE}\left(g.t\right)\)
\(BE\) : cạnh chung
\(\Rightarrow\Delta BED=\Delta BEC\left(đpcm\right)\)
c. Ta có:
\(\widehat{BED}+\widehat{DEI}=\widehat{BEC}+\widehat{CEI}\left(=180^o\right)\)
Mà \(\widehat{DEB}=\widehat{CEB}\left(\Delta BED=\Delta BEC\right)\)
\(\Rightarrow\widehat{DEI}=\widehat{CEI}\)
Xét \(\Delta IED\) và \(\Delta IEC\) có:
\(DE=CE\left(\Delta BED=\Delta BEC\right)\)
\(\widehat{DEI}=\widehat{CEI}\) (c.m trên)
\(EI\) : cạnh chung
\(\Rightarrow\Delta IED=\Delta IEC\left(đpcm\right)\)
Chúc bạn học tốt@@
a) ΔABC có:
\(\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^0\)
\(\Rightarrow\widehat{BAC}=180^0-\widehat{ABC}-\widehat{ACB}=180^0-60^0-40^0=80^0\)
b) Xét ΔBED và ΔBEC ta có:
BE: cạnh chung
\(\widehat{DBE}=\widehat{EBC}\left(GT\right)\)
BD = BC (GT)
=> ΔBED = ΔBEC (c - g - c)
c) ΔBED = ΔBEC (câu b)
=> ED = EC (2 cạnh tương ứng)
Và: \(\widehat{BED}=\widehat{BEC}\) (2 góc tương ứng)
Ta có: \(\widehat{BED}+\widehat{DEI}=180^0\) (kề bù)
\(\widehat{BEC}+\widehat{CEI}=180^0\)(kề bù)
Mà \(\widehat{BED}=\widehat{BEC}\) (cmt)
=> \(\widehat{DEI}=\widehat{CEI}\)
Xét ΔIED và ΔIEC ta có:
DE = EC (cmt)
\(\widehat{DEI}=\widehat{CEI}\) (cmt)
EI: cạnh chung
=> ΔIED = ΔIEC (c - g - c)
