hình tự vẽ
Ta có: \(\widehat{ADH}+\widehat{AEH}=90^o+90^o=180^o\)
suy ra ADHE nội tiếp
\(\Rightarrow\widehat{DEH}=\widehat{DAH}\Rightarrow\widehat{DEH}+\widehat{ABC}=90^o\)
do \(\widehat{DAH}+\widehat{ABC}=90^o\)
\(\Rightarrow\widehat{DBC}+\widehat{CED}=\widehat{CEH}+\widehat{DAH}+\widehat{DBC}=90^o+90^o=180^o\)
suy ra BDEC nội tiếp