Question

cứuu di 4h nộp rồi

Answer 1

Xét tg EFH và tg EGH, có:

\(\left\{{}\begin{matrix}\widehat{FEH}=\widehat{GEH}\left(gt\right)\\EH:canh\cdot chung\\\widehat{EHF}=\widehat{EHG}\left(=90^0\right)\end{matrix}\right.\)

\(=>\Delta EFH=\Delta EGH\left(g\cdot c\cdot g\right)\)

\(=>\widehat{F}=\widehat{G}\left(cgtu\right)\)

Answer 2

Xét ΔEHF và ΔEHG có:
\(\widehat{EHF}=\widehat{EHG}\left(=90^o\right)\)

Chung EH

\(\widehat{FEH}=\widehat{GEH}\left(gt\right)\)

⇒ΔEHF = ΔEHG (g.c.g)

⇒\(\widehat{F}=\widehat{G}\) (2 góc tương ứng)