Question

In the given figure, DE || BC, AE = 15cm, EC = 9cm, NC = 6 cm & BN = 24 cm. Find the value of ME & DM.

Answer 1

In $\Delta AME\&\Delta ANC$

$\angle A=\angle A$ (common)

$\angle AEM=\angle ACM$ (corresponding angle)

So $\Delta AME\sim \angle ANC$ (AA similarity)

$\frac{AM}{AN}=\frac{AE}{AC}=\frac{ME}{NC}$ (corresponding sides) ------- (1)

$\frac{AE}{AE+EC}=\frac{ME}{NC}\frac{15}{24}=\frac{ME}{6}$

ME = 3.75 cm

Now in $\Delta ADM\&\Delta ABN$

$\angle A=\angle A$ (common)

$\angle ADM=\angle ABN$ (corresponding angle)

$\frac{DM}{BN}=\frac{AM}{AN}$

but from (1) $\frac{AM}{AN}=\frac{AE}{AC}$

So $\frac{DM}{BN}=\frac{AE}{AC}$
$\frac{DM}{24}=\frac{15}{24}$
$\Rightarrow DM=15cm$