$MB=\dfrac{1}{2}MC$
$→MB=\dfrac{1}{3}BC$
Xét $ΔBDM$ và $ΔBAC$:
$\widehat{BDM}=\widehat{BAC}$ ($DM//AC$)
$\widehat{BMD}=\widehat{BCA}$ ($DM//AC$)
$→ΔBDM\simΔBAC$ mà $MB=\dfrac{1}{3}BC$
$→\dfrac{P_{ΔBDM}}{P_{ΔABC}}=\dfrac{1}{3}$
$→\dfrac{P_{ΔBDM}}{24}=\dfrac{1}{3}$
$→P_{ΔBDM}=8(cm)$
$MB=\dfrac{1}{2}MC$
$→MC=\dfrac{2}{3}BC$
Xét $ΔCME$ và $ΔCBA$:
$\widehat{CME}=\widehat{CBA}$ ($EM//AB$)
$\widehat{CEM}=\widehat{CAB}$ ($ME//AB$)
$→ΔCME\simΔCBA$ mà $MC=\dfrac{2}{3}BC$
$→\dfrac{P_{ΔCME}}{P_{ΔCBA}}=\dfrac{2}{3}$
$→\dfrac{P_{ΔCME}}{24}=\dfrac{2}{3}$
$→P_{ΔCME}=16(cm)$