Question

Answer 1

Bài 1:

2AB=3BC=5AC

=>\(\dfrac{2AB}{30}=\dfrac{3BC}{30}=\dfrac{5AC}{30}\)

=>\(\dfrac{AB}{15}=\dfrac{BC}{10}=\dfrac{AC}{6}\)

=>\(\dfrac{AB}{A'B'}=\dfrac{BC}{B'C'}=\dfrac{AC}{A'C'}\)

Xét ΔABC và ΔA'B'C' có

\(\dfrac{AB}{A'B'}=\dfrac{BC}{B'C'}=\dfrac{AC}{A'C'}\)

Do đó: ΔABC~ΔA'B'C'

Bài 2:

ΔEFG~ΔE'F'G'

=>\(\widehat{G}=\widehat{G'};\dfrac{EG}{E'G'}=\dfrac{FG}{F'G'}\)

\(\dfrac{EG}{E'G'}=\dfrac{FG}{F'G'}\)

=>\(\dfrac{EG}{E'G'}=\dfrac{2MG}{2NG'}=\dfrac{MG}{NG'}\)

Xét ΔEGM và ΔE'G'N có

\(\dfrac{EG}{E'G'}=\dfrac{GM}{G'N}\)

\(\widehat{G}=\widehat{G'}\)

Do đó: ΔEGM~ΔE'G'N