a,
Do \(AB.BD=BE.BC\Rightarrow\dfrac{AB}{BE}=\dfrac{BC}{BD}\)
Xét hai tam giác ABC và EBD có:
\(\left\{{}\begin{matrix}\dfrac{AB}{BE}=\dfrac{BC}{BD}\left(cht\right)\\\widehat{ABC}=\widehat{EBD}=90^0\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\Delta ABC\sim\Delta EBD\left(c.g.c\right)\)
b.
Do \(\Delta ABC\sim\Delta EBD\Rightarrow\widehat{ACB}=\widehat{EDB}\)
Lại có: \(\widehat{CEG}=\widehat{BED}\) (đối đỉnh)
Mà \(\left\{{}\begin{matrix}\widehat{ACB}+\widehat{CEG}+\widehat{CGE}=180^0\\\widehat{EDB}+\widehat{BED}+\widehat{EBD}=180^0\\\end{matrix}\right.\)
\(\Rightarrow\widehat{CGE}=\widehat{EBD}=90^0\)
\(\Rightarrow\Delta GEC\) vuông tại G
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