Question

In triangle $ABC$; $M$ is mid-point of $AB$, $N$ is mid-point of $AC$ and $D$ is any point in base $BC$. Use intercept theorem to show that $MN$ bisects $AD$.

Answer 1

In $△AMN$ and $△ABC$
$\angle MAN=\angle BAC$ .....(common)
$\frac{AM}{AB}=\frac{AN}{AC}=\frac{1}{2}$
$\therefore △AMN$ is similar to $△ABC$
$\therefore \angle AMN=\angle ABC⟶1$
In $△AMO$ and $△ABD$
$\Rightarrow \angle MAO=\angle BAD$ ....(common)
$\Rightarrow \angle AMO=\angle ABD$ .....(From equation 1)
$\Rightarrow AM=2AB$
Therefore, $△AMO$ and $△ABC$ are similar.
Thus $\frac{AM}{AB}=\frac{AO}{AD}$
$\Rightarrow \frac{AM}{2AM}=\frac{AO}{AD}$
$\Rightarrow AD=2AO$
$\Rightarrow AO=AD$
Therefore, $MN$ bisects $AD$
Hence, proved.