Question

In the given figure, $DE||BC,AE=15cm,EC=9cm,NC=6cm$ and $BN=24cm$.

ii) Find lengths of $ME$ and $DM$.

Answer 1

Step $1$ : Finding all possible pairs of similar triangles


Given: $DE||BC$

In $\Delta ADE$ and $\Delta ABC$,
$\angle ADE=\angle ABC$ [Corresponding angles]
$\angle AED=\angle ACB$ [Corresponding angles]
Therefore, by $AA$ criterion, $\Delta ADE$ ∼ $\Delta ABC$.

In $\Delta ADM$ and $\Delta ABN$,
$\angle ADM=\angle ABN$ [Corresponding angles]
$\angle DAM=\angle BAN$ [Common angle]
Therefore, by $AA$ criterion, $\Delta ADM$ ∼ $\Delta ABN$.
In $\Delta AME$ and $\Delta ANC$,
$\angle AME=\angle ANC$ [Corresponding angles]
$\angle MAE=\angle NAC$ [Common angle]
Therefore, by $AA$ criterion, $\Delta AME$ ∼ $\Delta ANC$.

So, all possible pairs of similar triangles are $\Delta ADM$ ∼ $\Delta ABN,\Delta AME$ ∼ $\Delta ANC$ and $\Delta ADE$ ∼ $\Delta ABC$.

Step $2$ : Find lengths of $ME$ and $DM$
Given: $AE=15cm,EC=9cm,$
NC=6 ~cm,BN=24 ~cm).

We know,
$\Delta ADM$ ∼ $\Delta ABN$,
$\Delta AME$ ∼ $\Delta ANC$,
$\Delta ADE$ ∼ $\Delta ABC$.

Therefore,
$\Rightarrow \frac{AD}{AB}=\frac{DM}{BN}=\frac{AM}{AN}\dots \left(I\right)$
$\Rightarrow \frac{AM}{AN}=\frac{ME}{NC}=\frac{AE}{AC}\dots \left(II\right)$
$\Rightarrow \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}\dots \left(III\right)$

On comparing $\left(I\right),\left(II\right)$ and $\left(III\right)$, we get;
$\Rightarrow \frac{ME}{NC}=\frac{DM}{BN}=\frac{AE}{AC}$
$\Rightarrow \frac{ME}{6}=\frac{DM}{24}=\frac{15}{15+9}[\text{Since},AC=AE+EC]$
$\Rightarrow \frac{ME}{6}=\frac{DM}{24}=\frac{15}{24}$

Therefore,
$\Rightarrow \frac{ME}{6}=\frac{15}{24}$
$\Rightarrow ME=6\times \frac{15}{24}$
$\Rightarrow ME=3.75cm$ and
$\Rightarrow \frac{DM}{24}=\frac{15}{24}$
$DM=24\times \frac{15}{9}$
$DM=15cm$

Hence, lengths of $ME$ and $DM$ are $3.75cm$ and $15cm$ respectively.