Giải thích các bước giải:
Vì \(BD\) là tia phân giác của \(\widehat{ABC}\) nên \(\widehat{ABD}=\widehat{CBD}=\frac{\widehat{ABC}}{2}\)
Vì \(ABC\) vuông tại \(A\) nên \(\widehat{A}=90^{\circ}\)
Xét \(\triangle ABC\) có \(\widehat{ABC}+\widehat{ACB}=90^{\circ}\) (tính chất tam giác vuông)
\(\Rightarrow \widehat{ABC}=90^{\circ}-\widehat{ACB}\)
\(\Rightarrow \frac{\widehat{ABC}}{2}=\frac{90^{\circ}-\widehat{ACB}}{2}\)
\(\Rightarrow \widehat{CBD}=45^{\circ}-\frac{\widehat{ACB}}{2}\)
Vì \(CH\perp DE\) nên \(\widehat{CHD}=90^{\circ}\)
Xét \(\triangle BHC\) có \(\widehat{HBC}+\widehat{BCH}=90^{\circ}\) (tính chất tam giác vuông)
\(\Rightarrow 45^{\circ}-\frac{\widehat{ACB}}{2}+\widehat{BCH}=90^{\circ}\)
\(\Rightarrow \widehat{BCH}-\frac{\widehat{ACB}}{2}=45^{\circ}\)
\(\Rightarrow \widehat{BCH}-\frac{\widehat{ACB}}{2}=\frac{\widehat{BCE}}{2}\) (Vì \(\widehat{BCE=90^{\circ}}\))
\(\Rightarrow \widehat{BCH}=\frac{\widehat{BCE}+\widehat{ACB}}{2}=\frac{2.\widehat{ACB}+\widehat{DCE}}{2}=\widehat{ACB}+\frac{\widehat{DCE}}{2}\)
\(\Rightarrow \widehat{BCH}-\widehat{ACB}=\frac{\widehat{DCE}}{2}\)
\(\Rightarrow \widehat{DCH}=\frac{\widehat{DCE}}{2}\)
\(\Rightarrow CH\) là tia phân giác \(\widehat{DCE}\)
