Gọi tia đối tia CB là Cx, tia đối tia BC là By, tia đối CD là CE, tia đối BD là BF
Xét \(\Delta\) ABC vuông tại A có:
\(\widehat{ABC}+\widehat{ACB}=180^o-\widehat{BAC}=180^o-90^o=90^o\)
Ta có:
\(\widehat{ACx}=180^o-\widehat{ACB}\)
\(\widehat{ABy}=180^o-\widehat{ABC}\)
\(\Rightarrow\widehat{ACx}+\widehat{ABy}=180^o-\widehat{ACB}+180^o-\widehat{ABC}=360^o-\left(\widehat{ACB}+\widehat{ABC}\right)=360^o-90^o=270^o\)
mà \(\widehat{ECx}=\dfrac{1}{2}\widehat{ACx}\) (CE là phân giác)
\(\widehat{FBy}=\dfrac{1}{2}\widehat{ABy}\) (BF là phân giác)
\(\Rightarrow\widehat{ECx}+\widehat{FBy}=\dfrac{1}{2}\left(\widehat{ACx}+\widehat{ABy}\right)=\dfrac{1}{2}\cdot270^o=135^o\)
lại có:
\(\widehat{ECx}=\widehat{DCB}\) (đối đỉnh)
\(\widehat{FBy}=\widehat{DBC}\) (đối đỉnh)
\(\Rightarrow\widehat{DCB}+\widehat{DBC}=135^o\)
Xét \(\Delta\) DBC có:
\(\widehat{BDC}=180^o-\left(\widehat{DCB}+\widehat{DBC}\right)=180^o-135^o=45^o\)
