In the given figure, DE || BC, AE = 15cm, EC = 9cm, NC = 6 cm & BN = 24 cm. The value of ME & DM are
3.75 & 15 cm respectively
In ΔAME & ΔANC
∠A=∠A (common)
∠AEM=∠ACM (corresponding angle)
So ΔAME∼∠ANC (AA similarity)
AMAN=AEAC=MENC (corresponding sides) ------- (1)
AEAE+EC=MENC1524=ME6
ME = 3.75 cm
Now in ΔADM & ΔABN
∠A=∠A (common)
∠ADM=∠ABN (corresponding angle)
DMBN=AMAN
but from (1) AMAN=AEAC
So DMBN=AEAC
DM24=1524
⇒ DM=15cm