Giải thích các bước giải:
Bài 1:
Ta có $(d)//AC\to MN//AC$
$\to \dfrac{MN}{AC}=\dfrac{BN}{BC}=\dfrac{BM}{BA}$
$\to \dfrac{3}{AC}=\dfrac{BN}{BC}=\dfrac{2}{5}$
$\to \dfrac{3}{AC}=\dfrac25\to AC=\dfrac{15}{2}$
Mà $\dfrac{BN}{BC}=\dfrac25$
$\to \dfrac{BC-CN}{CB}=\dfrac25$
$\to \dfrac{BC-4}{BC}=\dfrac25$
$\to 1-\dfrac4{BC}=\dfrac25$
$\to\dfrac4{CB}=\dfrac35$
$\to BC=\dfrac{20}{3}$
Bài 2:
Ta có $MN//AB//CD$
$\to \dfrac{MA}{MD}=\dfrac{QA}{QC}=\dfrac{NB}{NC}$
$\to \dfrac{10}{20}=\dfrac{QA}{35}=\dfrac{1}{NC}$
$\to \dfrac{1}{2}=\dfrac{QA}{35}=\dfrac{1}{NC}$
$\to AQ=\dfrac{35}{2}, NC=2$
