Question

Answer 1

\(a,\Delta BMD\) và \(\Delta CND\)có:

     \(\widehat{BMD}=\widehat{CND}\left(=90^o\right)\)

     \(\widehat{BDM}=\widehat{CDN}\) (2 góc đối đỉnh)

\(\Rightarrow\Delta BMD\sim\Delta CND\left(g-g\right)\)

Answer 2

b, \(\dfrac{BM}{CN}=\dfrac{BD}{CD}\) lại có AD là pg \(\dfrac{AB}{AC}=\dfrac{BD}{CD}=\dfrac{24}{28}=\dfrac{6}{7}\)

hay \(\dfrac{BM}{CN}=\dfrac{6}{7}\)