Chứng minh tam giác OBE = tam giác OCD biết tam giác ABC có AB=AC

Giải:
a) Vì \(\Delta ABC\) có AB = AC nên \(\Delta ABC\) cân tại A

\(\Rightarrow\widehat{B}=\widehat{C}\)

Xét \(\Delta EBC\) có: \(\widehat{B}+\widehat{C_1}=90^o\) ( do \(\widehat{BEO}=90^o\) )

Xét \(\Delta DBC\) có: \(\widehat{C}+\widehat{B_1}=90^o\) ( do \(\widehat{CDB}=90^o\) )

Mà \(\widehat{B}=\widehat{C}\Rightarrow\widehat{B_1}=\widehat{C_1}\) (*)

Xét \(\Delta EBC,\Delta DBC\) có:
\(\widehat{B}=\widehat{C}\)

\(BC\): cạnh chung

\(\widehat{B_1}=\widehat{C_1}\) ( theo (*) )

\(\Rightarrow\Delta EBC=\Delta DBC\left(g-c-g\right)\)

\(\Rightarrow BD=CE\) ( cạnh t/ứng ) ( đpcm )

\(\Rightarrow BE=CD\) ( cạnh t/ứng )

b) Ta có: \(\widehat{B}=\widehat{C}\)

\(\widehat{B_1}=\widehat{C_1}\)

\(\Rightarrow\widehat{B}-\widehat{B_1}=\widehat{C}-\widehat{C_1}\)

\(\Rightarrow\widehat{B_2}=\widehat{C_2}\) (**)

Xét \(\Delta OBE,\Delta OCD\) có:

\(\widehat{BEO}=\widehat{CDO}\left(=90^o\right)\)

BE = CD ( theo phần a )

\(\widehat{B_2}=\widehat{C_2}\) ( theo (**) )

\(\Rightarrow\Delta OBE=\Delta OCD\left(g-c-g\right)\) ( đpcm )