$a)$ Ta có:
$\dfrac{MA}{MB}=\dfrac{AN}{NC}=\dfrac{2}{3}$
$\widehat{A}$ là góc chung
$→ ΔMAN$ $ᔕ$ $ΔDAC$
$→ MN//BC$
Áp dụng pytago ta có:
$BC=\sqrt{3^2+4^2}=\sqrt{25}=5$
$→ \dfrac{MN}{DC}=\dfrac{AM}{AB}=\dfrac{2}{3}$
$→ MN=\dfrac{DC.2}{3}=\dfrac{10}{3}\ cm$
$b)$ Ta có:
$\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}$
$⇔ \dfrac{1}{AH^2}=\dfrac{1}{3^2}+\dfrac{1}{4^2}=\dfrac{25}{144}$
$⇒ AH=\dfrac{12}{5}$
$\dfrac{AK}{AH}=\dfrac{2}{3}$
$⇒ AK=\dfrac{12}{5}.\dfrac{2}{3}=\dfrac{24}{15}\ cm$