a) Xét \(\Delta ABE\) và \(\Delta ACF\) có:
\(\widehat{AEB}=\widehat{AFC}\)
\(\widehat{A}\) chung
\(\Rightarrow\Delta ABE\sim\Delta ACF\left(gn\right)\)
b) Vì \(\Delta ABE\sim\Delta ACF\)
\(\Rightarrow\widehat{ABE}=\widehat{ACF}\left(1\right)\)
Theo bài ra, ta có: AB // d
\(\Rightarrow\widehat{ABE}=\widehat{BED}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\widehat{ACF}=\widehat{BED}\)
Xét \(\Delta HED\) và \(\Delta HEC\) có:
\(\widehat{BED}=\widehat{ACF}\)
\(\widehat{EHC}\) chung
\(\Rightarrow\Delta HED\sim\Delta HEC\left(g-g\right)\)
\(\Rightarrow\dfrac{HE}{HD}=\dfrac{HC}{HE}\)
\(\Leftrightarrow HE^2=HD.HC\)