Question

Cho ΔABC, điểm O nằm trong tam giác. Các điểm M, N, P thuộc các traOA, OB, OC sao cho \(\dfrac{OA}{OM}\) = \(\dfrac{OB}{ON}\) =\(\dfrac{OC}{OP}\)  =\(\dfrac{3}{2}\). Cm ΔABC ∞ ΔMNP

Answer 1

Xét \(\Delta OAB\&\Delta OMN:\)

\(\dfrac{OA}{OM}=\dfrac{OB}{ON}=\dfrac{3}{2}\left(gt\right)\)

mà \(\widehat{OAB}=\widehat{OMN}\left(góc.chung\right)\)

\(\Rightarrow\Delta OAB\approx\Delta OMN\left(c.g.c\right)\)\(\Rightarrow\dfrac{AB}{MN}=\dfrac{OA}{OM}=\dfrac{3}{2}\left(1\right)\)

Chứng minh tương tự ta có :

\(\Delta OBC\approx\Delta ONP\left(c.g.c\right)\Rightarrow\dfrac{BC}{NP}=\dfrac{OP}{ON}=\dfrac{3}{2}\left(2\right)\)

\(\Delta OCA\approx\Delta OPM\left(c.g.c\right)\Rightarrow\dfrac{CA}{PM}=\dfrac{OC}{OP}=\dfrac{3}{2}\left(3\right)\)

\(\left(1\right);\left(2\right);\left(3\right)\Rightarrow\dfrac{AB}{MN}=\dfrac{BC}{NP}=\dfrac{CA}{PM}=\dfrac{3}{2}\)

\(\Rightarrow\Delta ABC\approx\Delta MNP\left(c.c.c\right)\)

\(\Rightarrowđpcm\)